The definition of continuity and functions defined at a single point
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In this question, the answers say that $lim_{x to p} f(x) = f(p) Longleftrightarrow f text{is continuous at} p$ fails if $f$ is only defined at a single point. Let us consider $f:{1} rightarrow mathbb{R}$. This is continuous at $1$, yet supposedly the standard limit definition fails.
I do not understand why. Write $D$ for the domain of $f$. Isn't the implication $forall epsilon >0 exists delta >0$ s.t $forall x in D$ satisfying $0<|x-1|<delta$ the inequality $|f(x) - f(1)|<epsilon$ holds true vacuously, since there is no $x in D$ satisfying $0<|x-1|<delta$ regardless of our choice of $delta$. Of course, if we define another function, say $g$, which has an isolated point but is defined elsewhere, then we simply choose $delta$ sufficiently small.
calculus limits continuity
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In this question, the answers say that $lim_{x to p} f(x) = f(p) Longleftrightarrow f text{is continuous at} p$ fails if $f$ is only defined at a single point. Let us consider $f:{1} rightarrow mathbb{R}$. This is continuous at $1$, yet supposedly the standard limit definition fails.
I do not understand why. Write $D$ for the domain of $f$. Isn't the implication $forall epsilon >0 exists delta >0$ s.t $forall x in D$ satisfying $0<|x-1|<delta$ the inequality $|f(x) - f(1)|<epsilon$ holds true vacuously, since there is no $x in D$ satisfying $0<|x-1|<delta$ regardless of our choice of $delta$. Of course, if we define another function, say $g$, which has an isolated point but is defined elsewhere, then we simply choose $delta$ sufficiently small.
calculus limits continuity
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add a comment |
$begingroup$
In this question, the answers say that $lim_{x to p} f(x) = f(p) Longleftrightarrow f text{is continuous at} p$ fails if $f$ is only defined at a single point. Let us consider $f:{1} rightarrow mathbb{R}$. This is continuous at $1$, yet supposedly the standard limit definition fails.
I do not understand why. Write $D$ for the domain of $f$. Isn't the implication $forall epsilon >0 exists delta >0$ s.t $forall x in D$ satisfying $0<|x-1|<delta$ the inequality $|f(x) - f(1)|<epsilon$ holds true vacuously, since there is no $x in D$ satisfying $0<|x-1|<delta$ regardless of our choice of $delta$. Of course, if we define another function, say $g$, which has an isolated point but is defined elsewhere, then we simply choose $delta$ sufficiently small.
calculus limits continuity
$endgroup$
In this question, the answers say that $lim_{x to p} f(x) = f(p) Longleftrightarrow f text{is continuous at} p$ fails if $f$ is only defined at a single point. Let us consider $f:{1} rightarrow mathbb{R}$. This is continuous at $1$, yet supposedly the standard limit definition fails.
I do not understand why. Write $D$ for the domain of $f$. Isn't the implication $forall epsilon >0 exists delta >0$ s.t $forall x in D$ satisfying $0<|x-1|<delta$ the inequality $|f(x) - f(1)|<epsilon$ holds true vacuously, since there is no $x in D$ satisfying $0<|x-1|<delta$ regardless of our choice of $delta$. Of course, if we define another function, say $g$, which has an isolated point but is defined elsewhere, then we simply choose $delta$ sufficiently small.
calculus limits continuity
calculus limits continuity
edited Apr 13 '17 at 12:21
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asked Mar 13 '16 at 0:47
MathematicsStudent1122MathematicsStudent1122
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3 Answers
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This depends on precisely how you phrase the definition of a limit. If you define the entire expression $$lim_{xto p}f(x)=L$$ as a single unit, then you are correct that if $p$ is an isolated point of the domain of $f$, this is true vacuously for every value of $L$. On the other hand, if you define $$lim_{xto p}f(x)$$ to refer to the (unique) number which satisfies the $epsilon$-$delta$ condition, then the limit does not exist if $p$ is isolated, since the definition is not satisfied by a unique number.
I don't know which definition is more commonly used in calculus books (or whether they even pick one version unambiguously), but I would consider the single unit definition to be the "correct" definition, since as you point out it eliminates the issue with continuity at isolated points.
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Most of the calculus books I have seen define it as a single unit and then subsequently prove uniqueness as a theorem with the implicit assumption that $p$ is not an isolated point. In general, would you say that the definition of continuity for real functions $lim_{x to p} f(x) = f(p)$ is technically precise? Or should the epsilon-delta def. be preferred?
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– MathematicsStudent1122
Mar 13 '16 at 15:30
1
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The single unit definition is better, as it matches the more general definition of continuity of maps between topological spaces.
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– Eric Wofsey
Mar 13 '16 at 16:49
add a comment |
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The answer is that when limits and continuity are correctly defined in a way that is applicable to this case, a function defined at a single point is always continuous. The typical calculus-book definition of limits needs to be revised, or else you can't generalize it. The difficulty is that the condition $0 < |x-1| < delta$ should really be replaced with $|x - 1| < delta$ for everything to work smoothly. The correct theoretical approach to relating limits and continuity is as follows.
Let $E$ and $F$ be sets. We will consider a function $f colon E_1 to F$, where $E_1$ is some subset of $E$. At the calculus level, $E$ and $F$ will be subsets of $mathbf{R}$, but in general they could be arbitrary metric spaces (or even topological spaces, although I won't discuss that case). For simplicity, I will write $|x - y|$ for the distance between $x$ and $y$, instead of $d(x,y)$ as I would in a metric space.
Let $A$ be a subset of $E_1$ and let $p in E$ be a point in the closure of $A$ (i.e., an "adherent point" of $A$). This means that for every $epsilon > 0$, there is some $a in A$ with $|a - p| < epsilon$. (If $p in A$, then this is always the case since we can take $a = p$.)
We define the notation $lim_{x to p, x in A} f(x) = L$ to mean that for every $epsilon > 0$, there exists $delta > 0$ such that whenever $x in A$ and $|x - p| < delta$, we have $|f(x) - L| < epsilon$. The assumption on $p$ is enough to imply the uniqueness of $L$.
For example, if $f colon mathbf{R} to mathbf{R}$, then the notation $lim_{x to 0, x in (0,+infty)} f(x)$ is simplified to $lim_{x to 0, x>0}f(x)$ or even, frequently, $lim_{x to 0^{+}} f(x)$. We can also have notations like $lim_{x to 0, text{$x$ rational}} f(x)$.
The notation $lim_{x to p} f(x)$ on its own should, in principle, mean $lim_{x to p, xin E_1} f(x)$, where $E_1$ is the entire domain of definition of $f$. In particular, if $p in E_1$, the existence of $lim_{x to p} f(x)$ automatically implies that this limit is equal to $f(p)$. We can then simply define $f$ to be continuous at a point $p in E_1$ whenever $lim_{x to p} f(x)$ exists.
But most often, in calculus books (or in more advanced books when the distinction is unimportant), $lim_{x to p} f(x)$ is taken to mean what we would technically write as $lim_{x to p, x ne p} f(x)$. The shorter notation is advantageous in some ways because limits of the form $lim_{x to p, x ne p} f(x)$ are considered so frequently.
However, from a theoretical perspective, considering the condition $x ne p$ to be implied is quite inconvenient. For example, it is this convention that makes "composition of limits" fail. That is, we can have $lim_{x to p, x ne p} f(x) = q$ and $lim_{y to q, y ne q} g(y) = r$ without necessarily having $lim_{x to p, x ne p} g(f(x)) = r$. (Take $f(x) = x sin 1/x$, $g(y) = 0$ for $y ne 0$, $g(0) = 1$, $p = q = 0$.) On the other hand, it is entirely true that if $f(A) subseteq B$, $lim_{x to p, x in A} f(x) = q$ and $lim_{y to q, y in B} g(y) = r$, then we must have $lim_{x to p, x in A} g(f(x)) = r$.
To come back to your original question, what happens if $E_1 = {p}$? In this case it is easy to see that the limit $lim_{x to p} f(x)$, which means (or ought to mean) $lim_{x to p, x in {p}} f(x)$, always exists and has the value $f(p)$. If you apply the calculus-book definition of a limit, it will denote $lim_{x to p, x in varnothing} f(x)$, which is not meaningful because $p$ is not an adherent point of $varnothing$.
Let me note in closing that the systematic use of notations like $lim_{x to p, x ne p} f(x)$ is not unheard of even at the calculus level. This approach was standard in high schools in France for many years, where $lim_{x to p} f(x)$ was defined the way I've defined it here. I haven't ever seen it done this way in English-language calculus textbooks, however.
Edit I've had a look at Apostol's analysis book, and the definition of a limit given there also assumes $x ne p$ implicitly. In that case $p$ is required to be an "accumulation point" of $A$, that is, an adherent point of $A - {p}$. One drawback to this approach (in addition to the composition issue I've already mentioned) is that he is required to give separate definitions of limits and continuity, rather than using limits to define continuity.
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If we allow $x = p$ wouldn't that have some weird consequences when considering otherwise continuous functions with a single removable discontinuity at $p$ which are still defined at $p$? With the definition you're proposing, the limit as $x rightarrow p$ would not exist.
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– MathematicsStudent1122
Mar 15 '16 at 22:13
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For example, take $f$ given by $$ f(x) = begin{cases} x & x neq 0 \ 1 & x=0 end{cases}$$ and take the limit as $x rightarrow 0$. If we allow $x=p$, the limit does not exist (taking $epsilon < 1$). However, clearly, $f$ can be made arbitrarily close to $0$ by making $x$ sufficiently close to $0$; indeed, this is the very motivation, to my knowledge, of the limit concept. Should the definition not formally encapsulate this motivation?
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– MathematicsStudent1122
Mar 15 '16 at 22:23
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According to the definitions I've stated, $lim_{x to 0} f(x)$ would not exist in that example, but we would have $lim_{x to 0, x ne 0} f(x) = 0$. The question is whether you can tolerate making the notation a bit longer so that the theory works more smoothly in other respects. (However, if the same $f$ were not defined at $0$, it would be fine to write $lim_{x to 0} f(x) = 0$.)
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– David
Mar 16 '16 at 4:25
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Topological Assessment
If $f: mathbb{R} to mathbb{R}$ is defined at a single point then its image must also be a single point, so that the function is defined as $f(x_0) = y_0$ for some $x_0, y_0 in mathbb{R}$.
I'm not sure about the answer and post this more for feedback. It seems that as a function $f: mathbb{R} to mathbb{R}$ it would not be continuous, but as a function $f:[x_0] to mathbb{R}$ then it would be.
A single point in $mathbb{R}$ is a closed set. If $O subset mathbb{R}$ is any open set that contains $y_0$ then its pre-image is the closed set $[x_0]$ which contardicts $f$ being continuous on $mathbb{R}$.
On the other hand, considered as a function from the domain $[x_0] subset mathbb{R}$ then $f:[x_0] to mathbb{R}$ would be continuous as $[x_0] $ is both closed and open in the subspace topology of $[x_0]$
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I agree with parts of what you wrote here, but when you write $f colon mathbf{R} to mathbf{R}$, that normally means that the domain of $f$ is all of $mathbf{R}$. If the domain of $f$ is ${x_0}$, you write $f colon {x_0} to mathbf{R}$. And in any event, if you want to apply the criterion on the preimage of an open set, you need to do it with respect to the subspace topology of the domain, not some larger space. If the domain of $f$ is a proper subset of $mathbf{R}$, it's not meaningful to discuss whether $f$ is continuous "considered as a function $f : mathbf{R} to mathbf{R}$."
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– David
Mar 16 '16 at 5:21
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@David Thanks. So, we would conclude that $f: [x_0] to mathbb R$ is continuous ?
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– Tom Collinge
Mar 16 '16 at 7:07
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Yes, more or less for the reason stated in your fourth paragraph. However, you also need to think about the possibility $O$ doesn't contain $y_0$. In that case, the inverse image of $O$ under $f$ is $varnothing$. Put more simply, if $O subseteq mathbf{R}$ is any open set (containing $y_0$ or not), then $f^{-1}(O) subseteq {x_0}$ is open because every subset of ${x_0}$ is open.
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– David
Mar 16 '16 at 20:44
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@David Thanks for feedback.
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– Tom Collinge
Mar 17 '16 at 9:09
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This depends on precisely how you phrase the definition of a limit. If you define the entire expression $$lim_{xto p}f(x)=L$$ as a single unit, then you are correct that if $p$ is an isolated point of the domain of $f$, this is true vacuously for every value of $L$. On the other hand, if you define $$lim_{xto p}f(x)$$ to refer to the (unique) number which satisfies the $epsilon$-$delta$ condition, then the limit does not exist if $p$ is isolated, since the definition is not satisfied by a unique number.
I don't know which definition is more commonly used in calculus books (or whether they even pick one version unambiguously), but I would consider the single unit definition to be the "correct" definition, since as you point out it eliminates the issue with continuity at isolated points.
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Most of the calculus books I have seen define it as a single unit and then subsequently prove uniqueness as a theorem with the implicit assumption that $p$ is not an isolated point. In general, would you say that the definition of continuity for real functions $lim_{x to p} f(x) = f(p)$ is technically precise? Or should the epsilon-delta def. be preferred?
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– MathematicsStudent1122
Mar 13 '16 at 15:30
1
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The single unit definition is better, as it matches the more general definition of continuity of maps between topological spaces.
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– Eric Wofsey
Mar 13 '16 at 16:49
add a comment |
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This depends on precisely how you phrase the definition of a limit. If you define the entire expression $$lim_{xto p}f(x)=L$$ as a single unit, then you are correct that if $p$ is an isolated point of the domain of $f$, this is true vacuously for every value of $L$. On the other hand, if you define $$lim_{xto p}f(x)$$ to refer to the (unique) number which satisfies the $epsilon$-$delta$ condition, then the limit does not exist if $p$ is isolated, since the definition is not satisfied by a unique number.
I don't know which definition is more commonly used in calculus books (or whether they even pick one version unambiguously), but I would consider the single unit definition to be the "correct" definition, since as you point out it eliminates the issue with continuity at isolated points.
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Most of the calculus books I have seen define it as a single unit and then subsequently prove uniqueness as a theorem with the implicit assumption that $p$ is not an isolated point. In general, would you say that the definition of continuity for real functions $lim_{x to p} f(x) = f(p)$ is technically precise? Or should the epsilon-delta def. be preferred?
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– MathematicsStudent1122
Mar 13 '16 at 15:30
1
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The single unit definition is better, as it matches the more general definition of continuity of maps between topological spaces.
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– Eric Wofsey
Mar 13 '16 at 16:49
add a comment |
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This depends on precisely how you phrase the definition of a limit. If you define the entire expression $$lim_{xto p}f(x)=L$$ as a single unit, then you are correct that if $p$ is an isolated point of the domain of $f$, this is true vacuously for every value of $L$. On the other hand, if you define $$lim_{xto p}f(x)$$ to refer to the (unique) number which satisfies the $epsilon$-$delta$ condition, then the limit does not exist if $p$ is isolated, since the definition is not satisfied by a unique number.
I don't know which definition is more commonly used in calculus books (or whether they even pick one version unambiguously), but I would consider the single unit definition to be the "correct" definition, since as you point out it eliminates the issue with continuity at isolated points.
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This depends on precisely how you phrase the definition of a limit. If you define the entire expression $$lim_{xto p}f(x)=L$$ as a single unit, then you are correct that if $p$ is an isolated point of the domain of $f$, this is true vacuously for every value of $L$. On the other hand, if you define $$lim_{xto p}f(x)$$ to refer to the (unique) number which satisfies the $epsilon$-$delta$ condition, then the limit does not exist if $p$ is isolated, since the definition is not satisfied by a unique number.
I don't know which definition is more commonly used in calculus books (or whether they even pick one version unambiguously), but I would consider the single unit definition to be the "correct" definition, since as you point out it eliminates the issue with continuity at isolated points.
edited Dec 4 '18 at 6:44
answered Mar 13 '16 at 1:06
Eric WofseyEric Wofsey
182k13209337
182k13209337
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Most of the calculus books I have seen define it as a single unit and then subsequently prove uniqueness as a theorem with the implicit assumption that $p$ is not an isolated point. In general, would you say that the definition of continuity for real functions $lim_{x to p} f(x) = f(p)$ is technically precise? Or should the epsilon-delta def. be preferred?
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– MathematicsStudent1122
Mar 13 '16 at 15:30
1
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The single unit definition is better, as it matches the more general definition of continuity of maps between topological spaces.
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– Eric Wofsey
Mar 13 '16 at 16:49
add a comment |
$begingroup$
Most of the calculus books I have seen define it as a single unit and then subsequently prove uniqueness as a theorem with the implicit assumption that $p$ is not an isolated point. In general, would you say that the definition of continuity for real functions $lim_{x to p} f(x) = f(p)$ is technically precise? Or should the epsilon-delta def. be preferred?
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– MathematicsStudent1122
Mar 13 '16 at 15:30
1
$begingroup$
The single unit definition is better, as it matches the more general definition of continuity of maps between topological spaces.
$endgroup$
– Eric Wofsey
Mar 13 '16 at 16:49
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Most of the calculus books I have seen define it as a single unit and then subsequently prove uniqueness as a theorem with the implicit assumption that $p$ is not an isolated point. In general, would you say that the definition of continuity for real functions $lim_{x to p} f(x) = f(p)$ is technically precise? Or should the epsilon-delta def. be preferred?
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– MathematicsStudent1122
Mar 13 '16 at 15:30
$begingroup$
Most of the calculus books I have seen define it as a single unit and then subsequently prove uniqueness as a theorem with the implicit assumption that $p$ is not an isolated point. In general, would you say that the definition of continuity for real functions $lim_{x to p} f(x) = f(p)$ is technically precise? Or should the epsilon-delta def. be preferred?
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– MathematicsStudent1122
Mar 13 '16 at 15:30
1
1
$begingroup$
The single unit definition is better, as it matches the more general definition of continuity of maps between topological spaces.
$endgroup$
– Eric Wofsey
Mar 13 '16 at 16:49
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The single unit definition is better, as it matches the more general definition of continuity of maps between topological spaces.
$endgroup$
– Eric Wofsey
Mar 13 '16 at 16:49
add a comment |
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The answer is that when limits and continuity are correctly defined in a way that is applicable to this case, a function defined at a single point is always continuous. The typical calculus-book definition of limits needs to be revised, or else you can't generalize it. The difficulty is that the condition $0 < |x-1| < delta$ should really be replaced with $|x - 1| < delta$ for everything to work smoothly. The correct theoretical approach to relating limits and continuity is as follows.
Let $E$ and $F$ be sets. We will consider a function $f colon E_1 to F$, where $E_1$ is some subset of $E$. At the calculus level, $E$ and $F$ will be subsets of $mathbf{R}$, but in general they could be arbitrary metric spaces (or even topological spaces, although I won't discuss that case). For simplicity, I will write $|x - y|$ for the distance between $x$ and $y$, instead of $d(x,y)$ as I would in a metric space.
Let $A$ be a subset of $E_1$ and let $p in E$ be a point in the closure of $A$ (i.e., an "adherent point" of $A$). This means that for every $epsilon > 0$, there is some $a in A$ with $|a - p| < epsilon$. (If $p in A$, then this is always the case since we can take $a = p$.)
We define the notation $lim_{x to p, x in A} f(x) = L$ to mean that for every $epsilon > 0$, there exists $delta > 0$ such that whenever $x in A$ and $|x - p| < delta$, we have $|f(x) - L| < epsilon$. The assumption on $p$ is enough to imply the uniqueness of $L$.
For example, if $f colon mathbf{R} to mathbf{R}$, then the notation $lim_{x to 0, x in (0,+infty)} f(x)$ is simplified to $lim_{x to 0, x>0}f(x)$ or even, frequently, $lim_{x to 0^{+}} f(x)$. We can also have notations like $lim_{x to 0, text{$x$ rational}} f(x)$.
The notation $lim_{x to p} f(x)$ on its own should, in principle, mean $lim_{x to p, xin E_1} f(x)$, where $E_1$ is the entire domain of definition of $f$. In particular, if $p in E_1$, the existence of $lim_{x to p} f(x)$ automatically implies that this limit is equal to $f(p)$. We can then simply define $f$ to be continuous at a point $p in E_1$ whenever $lim_{x to p} f(x)$ exists.
But most often, in calculus books (or in more advanced books when the distinction is unimportant), $lim_{x to p} f(x)$ is taken to mean what we would technically write as $lim_{x to p, x ne p} f(x)$. The shorter notation is advantageous in some ways because limits of the form $lim_{x to p, x ne p} f(x)$ are considered so frequently.
However, from a theoretical perspective, considering the condition $x ne p$ to be implied is quite inconvenient. For example, it is this convention that makes "composition of limits" fail. That is, we can have $lim_{x to p, x ne p} f(x) = q$ and $lim_{y to q, y ne q} g(y) = r$ without necessarily having $lim_{x to p, x ne p} g(f(x)) = r$. (Take $f(x) = x sin 1/x$, $g(y) = 0$ for $y ne 0$, $g(0) = 1$, $p = q = 0$.) On the other hand, it is entirely true that if $f(A) subseteq B$, $lim_{x to p, x in A} f(x) = q$ and $lim_{y to q, y in B} g(y) = r$, then we must have $lim_{x to p, x in A} g(f(x)) = r$.
To come back to your original question, what happens if $E_1 = {p}$? In this case it is easy to see that the limit $lim_{x to p} f(x)$, which means (or ought to mean) $lim_{x to p, x in {p}} f(x)$, always exists and has the value $f(p)$. If you apply the calculus-book definition of a limit, it will denote $lim_{x to p, x in varnothing} f(x)$, which is not meaningful because $p$ is not an adherent point of $varnothing$.
Let me note in closing that the systematic use of notations like $lim_{x to p, x ne p} f(x)$ is not unheard of even at the calculus level. This approach was standard in high schools in France for many years, where $lim_{x to p} f(x)$ was defined the way I've defined it here. I haven't ever seen it done this way in English-language calculus textbooks, however.
Edit I've had a look at Apostol's analysis book, and the definition of a limit given there also assumes $x ne p$ implicitly. In that case $p$ is required to be an "accumulation point" of $A$, that is, an adherent point of $A - {p}$. One drawback to this approach (in addition to the composition issue I've already mentioned) is that he is required to give separate definitions of limits and continuity, rather than using limits to define continuity.
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If we allow $x = p$ wouldn't that have some weird consequences when considering otherwise continuous functions with a single removable discontinuity at $p$ which are still defined at $p$? With the definition you're proposing, the limit as $x rightarrow p$ would not exist.
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– MathematicsStudent1122
Mar 15 '16 at 22:13
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For example, take $f$ given by $$ f(x) = begin{cases} x & x neq 0 \ 1 & x=0 end{cases}$$ and take the limit as $x rightarrow 0$. If we allow $x=p$, the limit does not exist (taking $epsilon < 1$). However, clearly, $f$ can be made arbitrarily close to $0$ by making $x$ sufficiently close to $0$; indeed, this is the very motivation, to my knowledge, of the limit concept. Should the definition not formally encapsulate this motivation?
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– MathematicsStudent1122
Mar 15 '16 at 22:23
1
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According to the definitions I've stated, $lim_{x to 0} f(x)$ would not exist in that example, but we would have $lim_{x to 0, x ne 0} f(x) = 0$. The question is whether you can tolerate making the notation a bit longer so that the theory works more smoothly in other respects. (However, if the same $f$ were not defined at $0$, it would be fine to write $lim_{x to 0} f(x) = 0$.)
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– David
Mar 16 '16 at 4:25
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The answer is that when limits and continuity are correctly defined in a way that is applicable to this case, a function defined at a single point is always continuous. The typical calculus-book definition of limits needs to be revised, or else you can't generalize it. The difficulty is that the condition $0 < |x-1| < delta$ should really be replaced with $|x - 1| < delta$ for everything to work smoothly. The correct theoretical approach to relating limits and continuity is as follows.
Let $E$ and $F$ be sets. We will consider a function $f colon E_1 to F$, where $E_1$ is some subset of $E$. At the calculus level, $E$ and $F$ will be subsets of $mathbf{R}$, but in general they could be arbitrary metric spaces (or even topological spaces, although I won't discuss that case). For simplicity, I will write $|x - y|$ for the distance between $x$ and $y$, instead of $d(x,y)$ as I would in a metric space.
Let $A$ be a subset of $E_1$ and let $p in E$ be a point in the closure of $A$ (i.e., an "adherent point" of $A$). This means that for every $epsilon > 0$, there is some $a in A$ with $|a - p| < epsilon$. (If $p in A$, then this is always the case since we can take $a = p$.)
We define the notation $lim_{x to p, x in A} f(x) = L$ to mean that for every $epsilon > 0$, there exists $delta > 0$ such that whenever $x in A$ and $|x - p| < delta$, we have $|f(x) - L| < epsilon$. The assumption on $p$ is enough to imply the uniqueness of $L$.
For example, if $f colon mathbf{R} to mathbf{R}$, then the notation $lim_{x to 0, x in (0,+infty)} f(x)$ is simplified to $lim_{x to 0, x>0}f(x)$ or even, frequently, $lim_{x to 0^{+}} f(x)$. We can also have notations like $lim_{x to 0, text{$x$ rational}} f(x)$.
The notation $lim_{x to p} f(x)$ on its own should, in principle, mean $lim_{x to p, xin E_1} f(x)$, where $E_1$ is the entire domain of definition of $f$. In particular, if $p in E_1$, the existence of $lim_{x to p} f(x)$ automatically implies that this limit is equal to $f(p)$. We can then simply define $f$ to be continuous at a point $p in E_1$ whenever $lim_{x to p} f(x)$ exists.
But most often, in calculus books (or in more advanced books when the distinction is unimportant), $lim_{x to p} f(x)$ is taken to mean what we would technically write as $lim_{x to p, x ne p} f(x)$. The shorter notation is advantageous in some ways because limits of the form $lim_{x to p, x ne p} f(x)$ are considered so frequently.
However, from a theoretical perspective, considering the condition $x ne p$ to be implied is quite inconvenient. For example, it is this convention that makes "composition of limits" fail. That is, we can have $lim_{x to p, x ne p} f(x) = q$ and $lim_{y to q, y ne q} g(y) = r$ without necessarily having $lim_{x to p, x ne p} g(f(x)) = r$. (Take $f(x) = x sin 1/x$, $g(y) = 0$ for $y ne 0$, $g(0) = 1$, $p = q = 0$.) On the other hand, it is entirely true that if $f(A) subseteq B$, $lim_{x to p, x in A} f(x) = q$ and $lim_{y to q, y in B} g(y) = r$, then we must have $lim_{x to p, x in A} g(f(x)) = r$.
To come back to your original question, what happens if $E_1 = {p}$? In this case it is easy to see that the limit $lim_{x to p} f(x)$, which means (or ought to mean) $lim_{x to p, x in {p}} f(x)$, always exists and has the value $f(p)$. If you apply the calculus-book definition of a limit, it will denote $lim_{x to p, x in varnothing} f(x)$, which is not meaningful because $p$ is not an adherent point of $varnothing$.
Let me note in closing that the systematic use of notations like $lim_{x to p, x ne p} f(x)$ is not unheard of even at the calculus level. This approach was standard in high schools in France for many years, where $lim_{x to p} f(x)$ was defined the way I've defined it here. I haven't ever seen it done this way in English-language calculus textbooks, however.
Edit I've had a look at Apostol's analysis book, and the definition of a limit given there also assumes $x ne p$ implicitly. In that case $p$ is required to be an "accumulation point" of $A$, that is, an adherent point of $A - {p}$. One drawback to this approach (in addition to the composition issue I've already mentioned) is that he is required to give separate definitions of limits and continuity, rather than using limits to define continuity.
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If we allow $x = p$ wouldn't that have some weird consequences when considering otherwise continuous functions with a single removable discontinuity at $p$ which are still defined at $p$? With the definition you're proposing, the limit as $x rightarrow p$ would not exist.
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– MathematicsStudent1122
Mar 15 '16 at 22:13
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For example, take $f$ given by $$ f(x) = begin{cases} x & x neq 0 \ 1 & x=0 end{cases}$$ and take the limit as $x rightarrow 0$. If we allow $x=p$, the limit does not exist (taking $epsilon < 1$). However, clearly, $f$ can be made arbitrarily close to $0$ by making $x$ sufficiently close to $0$; indeed, this is the very motivation, to my knowledge, of the limit concept. Should the definition not formally encapsulate this motivation?
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– MathematicsStudent1122
Mar 15 '16 at 22:23
1
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According to the definitions I've stated, $lim_{x to 0} f(x)$ would not exist in that example, but we would have $lim_{x to 0, x ne 0} f(x) = 0$. The question is whether you can tolerate making the notation a bit longer so that the theory works more smoothly in other respects. (However, if the same $f$ were not defined at $0$, it would be fine to write $lim_{x to 0} f(x) = 0$.)
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– David
Mar 16 '16 at 4:25
add a comment |
$begingroup$
The answer is that when limits and continuity are correctly defined in a way that is applicable to this case, a function defined at a single point is always continuous. The typical calculus-book definition of limits needs to be revised, or else you can't generalize it. The difficulty is that the condition $0 < |x-1| < delta$ should really be replaced with $|x - 1| < delta$ for everything to work smoothly. The correct theoretical approach to relating limits and continuity is as follows.
Let $E$ and $F$ be sets. We will consider a function $f colon E_1 to F$, where $E_1$ is some subset of $E$. At the calculus level, $E$ and $F$ will be subsets of $mathbf{R}$, but in general they could be arbitrary metric spaces (or even topological spaces, although I won't discuss that case). For simplicity, I will write $|x - y|$ for the distance between $x$ and $y$, instead of $d(x,y)$ as I would in a metric space.
Let $A$ be a subset of $E_1$ and let $p in E$ be a point in the closure of $A$ (i.e., an "adherent point" of $A$). This means that for every $epsilon > 0$, there is some $a in A$ with $|a - p| < epsilon$. (If $p in A$, then this is always the case since we can take $a = p$.)
We define the notation $lim_{x to p, x in A} f(x) = L$ to mean that for every $epsilon > 0$, there exists $delta > 0$ such that whenever $x in A$ and $|x - p| < delta$, we have $|f(x) - L| < epsilon$. The assumption on $p$ is enough to imply the uniqueness of $L$.
For example, if $f colon mathbf{R} to mathbf{R}$, then the notation $lim_{x to 0, x in (0,+infty)} f(x)$ is simplified to $lim_{x to 0, x>0}f(x)$ or even, frequently, $lim_{x to 0^{+}} f(x)$. We can also have notations like $lim_{x to 0, text{$x$ rational}} f(x)$.
The notation $lim_{x to p} f(x)$ on its own should, in principle, mean $lim_{x to p, xin E_1} f(x)$, where $E_1$ is the entire domain of definition of $f$. In particular, if $p in E_1$, the existence of $lim_{x to p} f(x)$ automatically implies that this limit is equal to $f(p)$. We can then simply define $f$ to be continuous at a point $p in E_1$ whenever $lim_{x to p} f(x)$ exists.
But most often, in calculus books (or in more advanced books when the distinction is unimportant), $lim_{x to p} f(x)$ is taken to mean what we would technically write as $lim_{x to p, x ne p} f(x)$. The shorter notation is advantageous in some ways because limits of the form $lim_{x to p, x ne p} f(x)$ are considered so frequently.
However, from a theoretical perspective, considering the condition $x ne p$ to be implied is quite inconvenient. For example, it is this convention that makes "composition of limits" fail. That is, we can have $lim_{x to p, x ne p} f(x) = q$ and $lim_{y to q, y ne q} g(y) = r$ without necessarily having $lim_{x to p, x ne p} g(f(x)) = r$. (Take $f(x) = x sin 1/x$, $g(y) = 0$ for $y ne 0$, $g(0) = 1$, $p = q = 0$.) On the other hand, it is entirely true that if $f(A) subseteq B$, $lim_{x to p, x in A} f(x) = q$ and $lim_{y to q, y in B} g(y) = r$, then we must have $lim_{x to p, x in A} g(f(x)) = r$.
To come back to your original question, what happens if $E_1 = {p}$? In this case it is easy to see that the limit $lim_{x to p} f(x)$, which means (or ought to mean) $lim_{x to p, x in {p}} f(x)$, always exists and has the value $f(p)$. If you apply the calculus-book definition of a limit, it will denote $lim_{x to p, x in varnothing} f(x)$, which is not meaningful because $p$ is not an adherent point of $varnothing$.
Let me note in closing that the systematic use of notations like $lim_{x to p, x ne p} f(x)$ is not unheard of even at the calculus level. This approach was standard in high schools in France for many years, where $lim_{x to p} f(x)$ was defined the way I've defined it here. I haven't ever seen it done this way in English-language calculus textbooks, however.
Edit I've had a look at Apostol's analysis book, and the definition of a limit given there also assumes $x ne p$ implicitly. In that case $p$ is required to be an "accumulation point" of $A$, that is, an adherent point of $A - {p}$. One drawback to this approach (in addition to the composition issue I've already mentioned) is that he is required to give separate definitions of limits and continuity, rather than using limits to define continuity.
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The answer is that when limits and continuity are correctly defined in a way that is applicable to this case, a function defined at a single point is always continuous. The typical calculus-book definition of limits needs to be revised, or else you can't generalize it. The difficulty is that the condition $0 < |x-1| < delta$ should really be replaced with $|x - 1| < delta$ for everything to work smoothly. The correct theoretical approach to relating limits and continuity is as follows.
Let $E$ and $F$ be sets. We will consider a function $f colon E_1 to F$, where $E_1$ is some subset of $E$. At the calculus level, $E$ and $F$ will be subsets of $mathbf{R}$, but in general they could be arbitrary metric spaces (or even topological spaces, although I won't discuss that case). For simplicity, I will write $|x - y|$ for the distance between $x$ and $y$, instead of $d(x,y)$ as I would in a metric space.
Let $A$ be a subset of $E_1$ and let $p in E$ be a point in the closure of $A$ (i.e., an "adherent point" of $A$). This means that for every $epsilon > 0$, there is some $a in A$ with $|a - p| < epsilon$. (If $p in A$, then this is always the case since we can take $a = p$.)
We define the notation $lim_{x to p, x in A} f(x) = L$ to mean that for every $epsilon > 0$, there exists $delta > 0$ such that whenever $x in A$ and $|x - p| < delta$, we have $|f(x) - L| < epsilon$. The assumption on $p$ is enough to imply the uniqueness of $L$.
For example, if $f colon mathbf{R} to mathbf{R}$, then the notation $lim_{x to 0, x in (0,+infty)} f(x)$ is simplified to $lim_{x to 0, x>0}f(x)$ or even, frequently, $lim_{x to 0^{+}} f(x)$. We can also have notations like $lim_{x to 0, text{$x$ rational}} f(x)$.
The notation $lim_{x to p} f(x)$ on its own should, in principle, mean $lim_{x to p, xin E_1} f(x)$, where $E_1$ is the entire domain of definition of $f$. In particular, if $p in E_1$, the existence of $lim_{x to p} f(x)$ automatically implies that this limit is equal to $f(p)$. We can then simply define $f$ to be continuous at a point $p in E_1$ whenever $lim_{x to p} f(x)$ exists.
But most often, in calculus books (or in more advanced books when the distinction is unimportant), $lim_{x to p} f(x)$ is taken to mean what we would technically write as $lim_{x to p, x ne p} f(x)$. The shorter notation is advantageous in some ways because limits of the form $lim_{x to p, x ne p} f(x)$ are considered so frequently.
However, from a theoretical perspective, considering the condition $x ne p$ to be implied is quite inconvenient. For example, it is this convention that makes "composition of limits" fail. That is, we can have $lim_{x to p, x ne p} f(x) = q$ and $lim_{y to q, y ne q} g(y) = r$ without necessarily having $lim_{x to p, x ne p} g(f(x)) = r$. (Take $f(x) = x sin 1/x$, $g(y) = 0$ for $y ne 0$, $g(0) = 1$, $p = q = 0$.) On the other hand, it is entirely true that if $f(A) subseteq B$, $lim_{x to p, x in A} f(x) = q$ and $lim_{y to q, y in B} g(y) = r$, then we must have $lim_{x to p, x in A} g(f(x)) = r$.
To come back to your original question, what happens if $E_1 = {p}$? In this case it is easy to see that the limit $lim_{x to p} f(x)$, which means (or ought to mean) $lim_{x to p, x in {p}} f(x)$, always exists and has the value $f(p)$. If you apply the calculus-book definition of a limit, it will denote $lim_{x to p, x in varnothing} f(x)$, which is not meaningful because $p$ is not an adherent point of $varnothing$.
Let me note in closing that the systematic use of notations like $lim_{x to p, x ne p} f(x)$ is not unheard of even at the calculus level. This approach was standard in high schools in France for many years, where $lim_{x to p} f(x)$ was defined the way I've defined it here. I haven't ever seen it done this way in English-language calculus textbooks, however.
Edit I've had a look at Apostol's analysis book, and the definition of a limit given there also assumes $x ne p$ implicitly. In that case $p$ is required to be an "accumulation point" of $A$, that is, an adherent point of $A - {p}$. One drawback to this approach (in addition to the composition issue I've already mentioned) is that he is required to give separate definitions of limits and continuity, rather than using limits to define continuity.
edited Mar 15 '16 at 5:35
answered Mar 15 '16 at 4:55
DavidDavid
5,223735
5,223735
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If we allow $x = p$ wouldn't that have some weird consequences when considering otherwise continuous functions with a single removable discontinuity at $p$ which are still defined at $p$? With the definition you're proposing, the limit as $x rightarrow p$ would not exist.
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– MathematicsStudent1122
Mar 15 '16 at 22:13
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For example, take $f$ given by $$ f(x) = begin{cases} x & x neq 0 \ 1 & x=0 end{cases}$$ and take the limit as $x rightarrow 0$. If we allow $x=p$, the limit does not exist (taking $epsilon < 1$). However, clearly, $f$ can be made arbitrarily close to $0$ by making $x$ sufficiently close to $0$; indeed, this is the very motivation, to my knowledge, of the limit concept. Should the definition not formally encapsulate this motivation?
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– MathematicsStudent1122
Mar 15 '16 at 22:23
1
$begingroup$
According to the definitions I've stated, $lim_{x to 0} f(x)$ would not exist in that example, but we would have $lim_{x to 0, x ne 0} f(x) = 0$. The question is whether you can tolerate making the notation a bit longer so that the theory works more smoothly in other respects. (However, if the same $f$ were not defined at $0$, it would be fine to write $lim_{x to 0} f(x) = 0$.)
$endgroup$
– David
Mar 16 '16 at 4:25
add a comment |
$begingroup$
If we allow $x = p$ wouldn't that have some weird consequences when considering otherwise continuous functions with a single removable discontinuity at $p$ which are still defined at $p$? With the definition you're proposing, the limit as $x rightarrow p$ would not exist.
$endgroup$
– MathematicsStudent1122
Mar 15 '16 at 22:13
$begingroup$
For example, take $f$ given by $$ f(x) = begin{cases} x & x neq 0 \ 1 & x=0 end{cases}$$ and take the limit as $x rightarrow 0$. If we allow $x=p$, the limit does not exist (taking $epsilon < 1$). However, clearly, $f$ can be made arbitrarily close to $0$ by making $x$ sufficiently close to $0$; indeed, this is the very motivation, to my knowledge, of the limit concept. Should the definition not formally encapsulate this motivation?
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– MathematicsStudent1122
Mar 15 '16 at 22:23
1
$begingroup$
According to the definitions I've stated, $lim_{x to 0} f(x)$ would not exist in that example, but we would have $lim_{x to 0, x ne 0} f(x) = 0$. The question is whether you can tolerate making the notation a bit longer so that the theory works more smoothly in other respects. (However, if the same $f$ were not defined at $0$, it would be fine to write $lim_{x to 0} f(x) = 0$.)
$endgroup$
– David
Mar 16 '16 at 4:25
$begingroup$
If we allow $x = p$ wouldn't that have some weird consequences when considering otherwise continuous functions with a single removable discontinuity at $p$ which are still defined at $p$? With the definition you're proposing, the limit as $x rightarrow p$ would not exist.
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– MathematicsStudent1122
Mar 15 '16 at 22:13
$begingroup$
If we allow $x = p$ wouldn't that have some weird consequences when considering otherwise continuous functions with a single removable discontinuity at $p$ which are still defined at $p$? With the definition you're proposing, the limit as $x rightarrow p$ would not exist.
$endgroup$
– MathematicsStudent1122
Mar 15 '16 at 22:13
$begingroup$
For example, take $f$ given by $$ f(x) = begin{cases} x & x neq 0 \ 1 & x=0 end{cases}$$ and take the limit as $x rightarrow 0$. If we allow $x=p$, the limit does not exist (taking $epsilon < 1$). However, clearly, $f$ can be made arbitrarily close to $0$ by making $x$ sufficiently close to $0$; indeed, this is the very motivation, to my knowledge, of the limit concept. Should the definition not formally encapsulate this motivation?
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– MathematicsStudent1122
Mar 15 '16 at 22:23
$begingroup$
For example, take $f$ given by $$ f(x) = begin{cases} x & x neq 0 \ 1 & x=0 end{cases}$$ and take the limit as $x rightarrow 0$. If we allow $x=p$, the limit does not exist (taking $epsilon < 1$). However, clearly, $f$ can be made arbitrarily close to $0$ by making $x$ sufficiently close to $0$; indeed, this is the very motivation, to my knowledge, of the limit concept. Should the definition not formally encapsulate this motivation?
$endgroup$
– MathematicsStudent1122
Mar 15 '16 at 22:23
1
1
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According to the definitions I've stated, $lim_{x to 0} f(x)$ would not exist in that example, but we would have $lim_{x to 0, x ne 0} f(x) = 0$. The question is whether you can tolerate making the notation a bit longer so that the theory works more smoothly in other respects. (However, if the same $f$ were not defined at $0$, it would be fine to write $lim_{x to 0} f(x) = 0$.)
$endgroup$
– David
Mar 16 '16 at 4:25
$begingroup$
According to the definitions I've stated, $lim_{x to 0} f(x)$ would not exist in that example, but we would have $lim_{x to 0, x ne 0} f(x) = 0$. The question is whether you can tolerate making the notation a bit longer so that the theory works more smoothly in other respects. (However, if the same $f$ were not defined at $0$, it would be fine to write $lim_{x to 0} f(x) = 0$.)
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– David
Mar 16 '16 at 4:25
add a comment |
$begingroup$
Topological Assessment
If $f: mathbb{R} to mathbb{R}$ is defined at a single point then its image must also be a single point, so that the function is defined as $f(x_0) = y_0$ for some $x_0, y_0 in mathbb{R}$.
I'm not sure about the answer and post this more for feedback. It seems that as a function $f: mathbb{R} to mathbb{R}$ it would not be continuous, but as a function $f:[x_0] to mathbb{R}$ then it would be.
A single point in $mathbb{R}$ is a closed set. If $O subset mathbb{R}$ is any open set that contains $y_0$ then its pre-image is the closed set $[x_0]$ which contardicts $f$ being continuous on $mathbb{R}$.
On the other hand, considered as a function from the domain $[x_0] subset mathbb{R}$ then $f:[x_0] to mathbb{R}$ would be continuous as $[x_0] $ is both closed and open in the subspace topology of $[x_0]$
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I agree with parts of what you wrote here, but when you write $f colon mathbf{R} to mathbf{R}$, that normally means that the domain of $f$ is all of $mathbf{R}$. If the domain of $f$ is ${x_0}$, you write $f colon {x_0} to mathbf{R}$. And in any event, if you want to apply the criterion on the preimage of an open set, you need to do it with respect to the subspace topology of the domain, not some larger space. If the domain of $f$ is a proper subset of $mathbf{R}$, it's not meaningful to discuss whether $f$ is continuous "considered as a function $f : mathbf{R} to mathbf{R}$."
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– David
Mar 16 '16 at 5:21
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@David Thanks. So, we would conclude that $f: [x_0] to mathbb R$ is continuous ?
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– Tom Collinge
Mar 16 '16 at 7:07
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Yes, more or less for the reason stated in your fourth paragraph. However, you also need to think about the possibility $O$ doesn't contain $y_0$. In that case, the inverse image of $O$ under $f$ is $varnothing$. Put more simply, if $O subseteq mathbf{R}$ is any open set (containing $y_0$ or not), then $f^{-1}(O) subseteq {x_0}$ is open because every subset of ${x_0}$ is open.
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– David
Mar 16 '16 at 20:44
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@David Thanks for feedback.
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– Tom Collinge
Mar 17 '16 at 9:09
add a comment |
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Topological Assessment
If $f: mathbb{R} to mathbb{R}$ is defined at a single point then its image must also be a single point, so that the function is defined as $f(x_0) = y_0$ for some $x_0, y_0 in mathbb{R}$.
I'm not sure about the answer and post this more for feedback. It seems that as a function $f: mathbb{R} to mathbb{R}$ it would not be continuous, but as a function $f:[x_0] to mathbb{R}$ then it would be.
A single point in $mathbb{R}$ is a closed set. If $O subset mathbb{R}$ is any open set that contains $y_0$ then its pre-image is the closed set $[x_0]$ which contardicts $f$ being continuous on $mathbb{R}$.
On the other hand, considered as a function from the domain $[x_0] subset mathbb{R}$ then $f:[x_0] to mathbb{R}$ would be continuous as $[x_0] $ is both closed and open in the subspace topology of $[x_0]$
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I agree with parts of what you wrote here, but when you write $f colon mathbf{R} to mathbf{R}$, that normally means that the domain of $f$ is all of $mathbf{R}$. If the domain of $f$ is ${x_0}$, you write $f colon {x_0} to mathbf{R}$. And in any event, if you want to apply the criterion on the preimage of an open set, you need to do it with respect to the subspace topology of the domain, not some larger space. If the domain of $f$ is a proper subset of $mathbf{R}$, it's not meaningful to discuss whether $f$ is continuous "considered as a function $f : mathbf{R} to mathbf{R}$."
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– David
Mar 16 '16 at 5:21
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@David Thanks. So, we would conclude that $f: [x_0] to mathbb R$ is continuous ?
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– Tom Collinge
Mar 16 '16 at 7:07
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Yes, more or less for the reason stated in your fourth paragraph. However, you also need to think about the possibility $O$ doesn't contain $y_0$. In that case, the inverse image of $O$ under $f$ is $varnothing$. Put more simply, if $O subseteq mathbf{R}$ is any open set (containing $y_0$ or not), then $f^{-1}(O) subseteq {x_0}$ is open because every subset of ${x_0}$ is open.
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– David
Mar 16 '16 at 20:44
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@David Thanks for feedback.
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– Tom Collinge
Mar 17 '16 at 9:09
add a comment |
$begingroup$
Topological Assessment
If $f: mathbb{R} to mathbb{R}$ is defined at a single point then its image must also be a single point, so that the function is defined as $f(x_0) = y_0$ for some $x_0, y_0 in mathbb{R}$.
I'm not sure about the answer and post this more for feedback. It seems that as a function $f: mathbb{R} to mathbb{R}$ it would not be continuous, but as a function $f:[x_0] to mathbb{R}$ then it would be.
A single point in $mathbb{R}$ is a closed set. If $O subset mathbb{R}$ is any open set that contains $y_0$ then its pre-image is the closed set $[x_0]$ which contardicts $f$ being continuous on $mathbb{R}$.
On the other hand, considered as a function from the domain $[x_0] subset mathbb{R}$ then $f:[x_0] to mathbb{R}$ would be continuous as $[x_0] $ is both closed and open in the subspace topology of $[x_0]$
$endgroup$
Topological Assessment
If $f: mathbb{R} to mathbb{R}$ is defined at a single point then its image must also be a single point, so that the function is defined as $f(x_0) = y_0$ for some $x_0, y_0 in mathbb{R}$.
I'm not sure about the answer and post this more for feedback. It seems that as a function $f: mathbb{R} to mathbb{R}$ it would not be continuous, but as a function $f:[x_0] to mathbb{R}$ then it would be.
A single point in $mathbb{R}$ is a closed set. If $O subset mathbb{R}$ is any open set that contains $y_0$ then its pre-image is the closed set $[x_0]$ which contardicts $f$ being continuous on $mathbb{R}$.
On the other hand, considered as a function from the domain $[x_0] subset mathbb{R}$ then $f:[x_0] to mathbb{R}$ would be continuous as $[x_0] $ is both closed and open in the subspace topology of $[x_0]$
edited Mar 15 '16 at 7:40
answered Mar 15 '16 at 7:33
Tom CollingeTom Collinge
4,5491133
4,5491133
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I agree with parts of what you wrote here, but when you write $f colon mathbf{R} to mathbf{R}$, that normally means that the domain of $f$ is all of $mathbf{R}$. If the domain of $f$ is ${x_0}$, you write $f colon {x_0} to mathbf{R}$. And in any event, if you want to apply the criterion on the preimage of an open set, you need to do it with respect to the subspace topology of the domain, not some larger space. If the domain of $f$ is a proper subset of $mathbf{R}$, it's not meaningful to discuss whether $f$ is continuous "considered as a function $f : mathbf{R} to mathbf{R}$."
$endgroup$
– David
Mar 16 '16 at 5:21
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@David Thanks. So, we would conclude that $f: [x_0] to mathbb R$ is continuous ?
$endgroup$
– Tom Collinge
Mar 16 '16 at 7:07
$begingroup$
Yes, more or less for the reason stated in your fourth paragraph. However, you also need to think about the possibility $O$ doesn't contain $y_0$. In that case, the inverse image of $O$ under $f$ is $varnothing$. Put more simply, if $O subseteq mathbf{R}$ is any open set (containing $y_0$ or not), then $f^{-1}(O) subseteq {x_0}$ is open because every subset of ${x_0}$ is open.
$endgroup$
– David
Mar 16 '16 at 20:44
$begingroup$
@David Thanks for feedback.
$endgroup$
– Tom Collinge
Mar 17 '16 at 9:09
add a comment |
$begingroup$
I agree with parts of what you wrote here, but when you write $f colon mathbf{R} to mathbf{R}$, that normally means that the domain of $f$ is all of $mathbf{R}$. If the domain of $f$ is ${x_0}$, you write $f colon {x_0} to mathbf{R}$. And in any event, if you want to apply the criterion on the preimage of an open set, you need to do it with respect to the subspace topology of the domain, not some larger space. If the domain of $f$ is a proper subset of $mathbf{R}$, it's not meaningful to discuss whether $f$ is continuous "considered as a function $f : mathbf{R} to mathbf{R}$."
$endgroup$
– David
Mar 16 '16 at 5:21
$begingroup$
@David Thanks. So, we would conclude that $f: [x_0] to mathbb R$ is continuous ?
$endgroup$
– Tom Collinge
Mar 16 '16 at 7:07
$begingroup$
Yes, more or less for the reason stated in your fourth paragraph. However, you also need to think about the possibility $O$ doesn't contain $y_0$. In that case, the inverse image of $O$ under $f$ is $varnothing$. Put more simply, if $O subseteq mathbf{R}$ is any open set (containing $y_0$ or not), then $f^{-1}(O) subseteq {x_0}$ is open because every subset of ${x_0}$ is open.
$endgroup$
– David
Mar 16 '16 at 20:44
$begingroup$
@David Thanks for feedback.
$endgroup$
– Tom Collinge
Mar 17 '16 at 9:09
$begingroup$
I agree with parts of what you wrote here, but when you write $f colon mathbf{R} to mathbf{R}$, that normally means that the domain of $f$ is all of $mathbf{R}$. If the domain of $f$ is ${x_0}$, you write $f colon {x_0} to mathbf{R}$. And in any event, if you want to apply the criterion on the preimage of an open set, you need to do it with respect to the subspace topology of the domain, not some larger space. If the domain of $f$ is a proper subset of $mathbf{R}$, it's not meaningful to discuss whether $f$ is continuous "considered as a function $f : mathbf{R} to mathbf{R}$."
$endgroup$
– David
Mar 16 '16 at 5:21
$begingroup$
I agree with parts of what you wrote here, but when you write $f colon mathbf{R} to mathbf{R}$, that normally means that the domain of $f$ is all of $mathbf{R}$. If the domain of $f$ is ${x_0}$, you write $f colon {x_0} to mathbf{R}$. And in any event, if you want to apply the criterion on the preimage of an open set, you need to do it with respect to the subspace topology of the domain, not some larger space. If the domain of $f$ is a proper subset of $mathbf{R}$, it's not meaningful to discuss whether $f$ is continuous "considered as a function $f : mathbf{R} to mathbf{R}$."
$endgroup$
– David
Mar 16 '16 at 5:21
$begingroup$
@David Thanks. So, we would conclude that $f: [x_0] to mathbb R$ is continuous ?
$endgroup$
– Tom Collinge
Mar 16 '16 at 7:07
$begingroup$
@David Thanks. So, we would conclude that $f: [x_0] to mathbb R$ is continuous ?
$endgroup$
– Tom Collinge
Mar 16 '16 at 7:07
$begingroup$
Yes, more or less for the reason stated in your fourth paragraph. However, you also need to think about the possibility $O$ doesn't contain $y_0$. In that case, the inverse image of $O$ under $f$ is $varnothing$. Put more simply, if $O subseteq mathbf{R}$ is any open set (containing $y_0$ or not), then $f^{-1}(O) subseteq {x_0}$ is open because every subset of ${x_0}$ is open.
$endgroup$
– David
Mar 16 '16 at 20:44
$begingroup$
Yes, more or less for the reason stated in your fourth paragraph. However, you also need to think about the possibility $O$ doesn't contain $y_0$. In that case, the inverse image of $O$ under $f$ is $varnothing$. Put more simply, if $O subseteq mathbf{R}$ is any open set (containing $y_0$ or not), then $f^{-1}(O) subseteq {x_0}$ is open because every subset of ${x_0}$ is open.
$endgroup$
– David
Mar 16 '16 at 20:44
$begingroup$
@David Thanks for feedback.
$endgroup$
– Tom Collinge
Mar 17 '16 at 9:09
$begingroup$
@David Thanks for feedback.
$endgroup$
– Tom Collinge
Mar 17 '16 at 9:09
add a comment |
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