The set of points $p$ such that $|p-p_0|>r$ is open, for any $p_0$ and any $r ge 0$.
$begingroup$
I started the problem off by assuming that $|p-p_0| = r + delta$
and that $|q-p| < frac{delta}{2}$ or equivalently that $q in B(p, frac{delta}{2})$.
Then we have that
$|q-p_0|=|(p-p_0) + (q-p)| le |p-p_0| + |q-p| < r + delta + frac{delta}{2}$.
Then,
$r < |q-p_0| < r + frac{3 delta}{2}$. Subtracting $r$ from the inequality I get that
$0<|q-p_0|-r< frac{3 delta}{2}$. But I'm not sure that I can do this since I think that I'm supposed to show that $r< |q-p_0|$ which I get from $0<|q-p_0| - r$. Although, I already assumed that $r<|q-p_0|$ so things seem off, but I don't know how to fix this proof.
real-analysis
asked Dec 9 '18 at 19:12
K.MK.M
693412
$endgroup$
add a comment |
$begingroup$
I started the problem off by assuming that $|p-p_0| = r + delta$
and that $|q-p| < frac{delta}{2}$ or equivalently that $q in B(p, frac{delta}{2})$.
Then we have that
$|q-p_0|=|(p-p_0) + (q-p)| le |p-p_0| + |q-p| < r + delta + frac{delta}{2}$.
Then,
$r < |q-p_0| < r + frac{3 delta}{2}$. Subtracting $r$ from the inequality I get that
$0<|q-p_0|-r< frac{3 delta}{2}$. But I'm not sure that I can do this since I think that I'm supposed to show that $r< |q-p_0|$ which I get from $0<|q-p_0| - r$. Although, I already assumed that $r<|q-p_0|$ so things seem off, but I don't know how to fix this proof.
real-analysis
asked Dec 9 '18 at 19:12
K.MK.M
693412
$endgroup$
add a comment |
$begingroup$
I started the problem off by assuming that $|p-p_0| = r + delta$
and that $|q-p| < frac{delta}{2}$ or equivalently that $q in B(p, frac{delta}{2})$.
Then we have that
$|q-p_0|=|(p-p_0) + (q-p)| le |p-p_0| + |q-p| < r + delta + frac{delta}{2}$.
Then,
$r < |q-p_0| < r + frac{3 delta}{2}$. Subtracting $r$ from the inequality I get that
$0<|q-p_0|-r< frac{3 delta}{2}$. But I'm not sure that I can do this since I think that I'm supposed to show that $r< |q-p_0|$ which I get from $0<|q-p_0| - r$. Although, I already assumed that $r<|q-p_0|$ so things seem off, but I don't know how to fix this proof.
real-analysis
asked Dec 9 '18 at 19:12
K.MK.M
693412
$endgroup$
I started the problem off by assuming that $|p-p_0| = r + delta$
and that $|q-p| < frac{delta}{2}$ or equivalently that $q in B(p, frac{delta}{2})$.
Then we have that
$|q-p_0|=|(p-p_0) + (q-p)| le |p-p_0| + |q-p| < r + delta + frac{delta}{2}$.
Then,
$r < |q-p_0| < r + frac{3 delta}{2}$. Subtracting $r$ from the inequality I get that
$0<|q-p_0|-r< frac{3 delta}{2}$. But I'm not sure that I can do this since I think that I'm supposed to show that $r< |q-p_0|$ which I get from $0<|q-p_0| - r$. Although, I already assumed that $r<|q-p_0|$ so things seem off, but I don't know how to fix this proof.
real-analysis
real-analysis
asked Dec 9 '18 at 19:12
K.MK.M
693412
asked Dec 9 '18 at 19:12
K.MK.M
693412
asked Dec 9 '18 at 19:12
K.MK.M
693412
asked Dec 9 '18 at 19:12
K.MK.M
693412
asked Dec 9 '18 at 19:12
K.MK.M
693412
693412
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
How much topology do you know? Depending on what level of exposure you have had you may be able to resolve this quite easily.
My approach would be something like the following:
Let $p_0inmathbb{R}$ be fixed, let $rgeq 0$, and consider the set of points ${p : |p-p_0|>r}$. This set is equal to ${p : p_0-p>r}cup{p : p-p_0>r} = (-infty,p_0-r)cup(p_0+r,infty)$. The union of two open sets is open, and what was needed to be shown has been.
As far as your own proof is concerned, you basically just got a little too obsessed with your inequalities; all you should be trying to show is that there is an open neighborhood surrounding $p$ entirely contained inside your set. If you work with a point $p$ that is $r+delta$ away from $p_0$, then the ball you have defined will suffice, but you won't show that via the triangle inequality. You will still probably want to break into cases.
answered Dec 9 '18 at 19:19
GenericMathematicianGenericMathematician
863
$endgroup$
$begingroup$
I know very basic topology. When you have the interval $(-infty, p_0-r)$, is that because $p_0-r>p$, do we just assume that $p> -infty$? Also with the cases that you mentioned, do you mean when $q$ is in the ball surrounding $p$ and contained in the "line" connecting $p$ to $p_0$ and when $q$ is not on the "line" connecting $p_0$ to $p$?
$endgroup$
– K.M
Dec 9 '18 at 19:34
1
$begingroup$
@K.M We don't have to assume that $p>-infty$; $p$ is a real number, so it has to be larger than $-infty$. The cases I am referring to are when $p>p_0$ and $p<p_0$. This will allow you to get rid of those pesky absolute value bars.
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:37
$begingroup$
I thought that $p$ and $p_0$ are points, do $p$ and $p_0$ denote the distances from $0$, respectively?
$endgroup$
– K.M
Dec 9 '18 at 19:43
$begingroup$
@K.M They are both specific points, but they are points in $mathbb{R}$, so we can compare them, add them, subtract them, etc. There is no difference between a point and a signed distance from $0$ in $mathbb{R}$.
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:46
1
$begingroup$
@K.M If we were working in $mathbb{R}^n$ then "absolute values" really mean euclidean distances, and then you can't break into cases because there is no finite number of directions to consider. You would then want to be extra careful in your analysis, either leveraging more topology or cleverly arguing with the reverse triangle inequality.
$endgroup$
– GenericMathematician
Dec 9 '18 at 20:01
|
show 1 more comment
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1 Answer
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$begingroup$
How much topology do you know? Depending on what level of exposure you have had you may be able to resolve this quite easily.
My approach would be something like the following:
Let $p_0inmathbb{R}$ be fixed, let $rgeq 0$, and consider the set of points ${p : |p-p_0|>r}$. This set is equal to ${p : p_0-p>r}cup{p : p-p_0>r} = (-infty,p_0-r)cup(p_0+r,infty)$. The union of two open sets is open, and what was needed to be shown has been.
As far as your own proof is concerned, you basically just got a little too obsessed with your inequalities; all you should be trying to show is that there is an open neighborhood surrounding $p$ entirely contained inside your set. If you work with a point $p$ that is $r+delta$ away from $p_0$, then the ball you have defined will suffice, but you won't show that via the triangle inequality. You will still probably want to break into cases.
answered Dec 9 '18 at 19:19
GenericMathematicianGenericMathematician
863
$endgroup$
$begingroup$
I know very basic topology. When you have the interval $(-infty, p_0-r)$, is that because $p_0-r>p$, do we just assume that $p> -infty$? Also with the cases that you mentioned, do you mean when $q$ is in the ball surrounding $p$ and contained in the "line" connecting $p$ to $p_0$ and when $q$ is not on the "line" connecting $p_0$ to $p$?
$endgroup$
– K.M
Dec 9 '18 at 19:34
1
$begingroup$
@K.M We don't have to assume that $p>-infty$; $p$ is a real number, so it has to be larger than $-infty$. The cases I am referring to are when $p>p_0$ and $p<p_0$. This will allow you to get rid of those pesky absolute value bars.
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:37
$begingroup$
I thought that $p$ and $p_0$ are points, do $p$ and $p_0$ denote the distances from $0$, respectively?
$endgroup$
– K.M
Dec 9 '18 at 19:43
$begingroup$
@K.M They are both specific points, but they are points in $mathbb{R}$, so we can compare them, add them, subtract them, etc. There is no difference between a point and a signed distance from $0$ in $mathbb{R}$.
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:46
1
$begingroup$
@K.M If we were working in $mathbb{R}^n$ then "absolute values" really mean euclidean distances, and then you can't break into cases because there is no finite number of directions to consider. You would then want to be extra careful in your analysis, either leveraging more topology or cleverly arguing with the reverse triangle inequality.
$endgroup$
– GenericMathematician
Dec 9 '18 at 20:01
|
show 1 more comment
$begingroup$
How much topology do you know? Depending on what level of exposure you have had you may be able to resolve this quite easily.
My approach would be something like the following:
Let $p_0inmathbb{R}$ be fixed, let $rgeq 0$, and consider the set of points ${p : |p-p_0|>r}$. This set is equal to ${p : p_0-p>r}cup{p : p-p_0>r} = (-infty,p_0-r)cup(p_0+r,infty)$. The union of two open sets is open, and what was needed to be shown has been.
As far as your own proof is concerned, you basically just got a little too obsessed with your inequalities; all you should be trying to show is that there is an open neighborhood surrounding $p$ entirely contained inside your set. If you work with a point $p$ that is $r+delta$ away from $p_0$, then the ball you have defined will suffice, but you won't show that via the triangle inequality. You will still probably want to break into cases.
answered Dec 9 '18 at 19:19
GenericMathematicianGenericMathematician
863
$endgroup$
$begingroup$
I know very basic topology. When you have the interval $(-infty, p_0-r)$, is that because $p_0-r>p$, do we just assume that $p> -infty$? Also with the cases that you mentioned, do you mean when $q$ is in the ball surrounding $p$ and contained in the "line" connecting $p$ to $p_0$ and when $q$ is not on the "line" connecting $p_0$ to $p$?
$endgroup$
– K.M
Dec 9 '18 at 19:34
1
$begingroup$
@K.M We don't have to assume that $p>-infty$; $p$ is a real number, so it has to be larger than $-infty$. The cases I am referring to are when $p>p_0$ and $p<p_0$. This will allow you to get rid of those pesky absolute value bars.
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:37
$begingroup$
I thought that $p$ and $p_0$ are points, do $p$ and $p_0$ denote the distances from $0$, respectively?
$endgroup$
– K.M
Dec 9 '18 at 19:43
$begingroup$
@K.M They are both specific points, but they are points in $mathbb{R}$, so we can compare them, add them, subtract them, etc. There is no difference between a point and a signed distance from $0$ in $mathbb{R}$.
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:46
1
$begingroup$
@K.M If we were working in $mathbb{R}^n$ then "absolute values" really mean euclidean distances, and then you can't break into cases because there is no finite number of directions to consider. You would then want to be extra careful in your analysis, either leveraging more topology or cleverly arguing with the reverse triangle inequality.
$endgroup$
– GenericMathematician
Dec 9 '18 at 20:01
|
show 1 more comment
$begingroup$
How much topology do you know? Depending on what level of exposure you have had you may be able to resolve this quite easily.
My approach would be something like the following:
Let $p_0inmathbb{R}$ be fixed, let $rgeq 0$, and consider the set of points ${p : |p-p_0|>r}$. This set is equal to ${p : p_0-p>r}cup{p : p-p_0>r} = (-infty,p_0-r)cup(p_0+r,infty)$. The union of two open sets is open, and what was needed to be shown has been.
As far as your own proof is concerned, you basically just got a little too obsessed with your inequalities; all you should be trying to show is that there is an open neighborhood surrounding $p$ entirely contained inside your set. If you work with a point $p$ that is $r+delta$ away from $p_0$, then the ball you have defined will suffice, but you won't show that via the triangle inequality. You will still probably want to break into cases.
answered Dec 9 '18 at 19:19
GenericMathematicianGenericMathematician
863
$endgroup$
How much topology do you know? Depending on what level of exposure you have had you may be able to resolve this quite easily.
My approach would be something like the following:
Let $p_0inmathbb{R}$ be fixed, let $rgeq 0$, and consider the set of points ${p : |p-p_0|>r}$. This set is equal to ${p : p_0-p>r}cup{p : p-p_0>r} = (-infty,p_0-r)cup(p_0+r,infty)$. The union of two open sets is open, and what was needed to be shown has been.
As far as your own proof is concerned, you basically just got a little too obsessed with your inequalities; all you should be trying to show is that there is an open neighborhood surrounding $p$ entirely contained inside your set. If you work with a point $p$ that is $r+delta$ away from $p_0$, then the ball you have defined will suffice, but you won't show that via the triangle inequality. You will still probably want to break into cases.
answered Dec 9 '18 at 19:19
GenericMathematicianGenericMathematician
863
edited Dec 9 '18 at 19:24
answered Dec 9 '18 at 19:19
GenericMathematicianGenericMathematician
863
answered Dec 9 '18 at 19:19
GenericMathematicianGenericMathematician
863
answered Dec 9 '18 at 19:19
GenericMathematicianGenericMathematician
863
863
$begingroup$
I know very basic topology. When you have the interval $(-infty, p_0-r)$, is that because $p_0-r>p$, do we just assume that $p> -infty$? Also with the cases that you mentioned, do you mean when $q$ is in the ball surrounding $p$ and contained in the "line" connecting $p$ to $p_0$ and when $q$ is not on the "line" connecting $p_0$ to $p$?
$endgroup$
– K.M
Dec 9 '18 at 19:34
1
$begingroup$
@K.M We don't have to assume that $p>-infty$; $p$ is a real number, so it has to be larger than $-infty$. The cases I am referring to are when $p>p_0$ and $p<p_0$. This will allow you to get rid of those pesky absolute value bars.
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:37
$begingroup$
I thought that $p$ and $p_0$ are points, do $p$ and $p_0$ denote the distances from $0$, respectively?
$endgroup$
– K.M
Dec 9 '18 at 19:43
$begingroup$
@K.M They are both specific points, but they are points in $mathbb{R}$, so we can compare them, add them, subtract them, etc. There is no difference between a point and a signed distance from $0$ in $mathbb{R}$.
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:46
1
$begingroup$
@K.M If we were working in $mathbb{R}^n$ then "absolute values" really mean euclidean distances, and then you can't break into cases because there is no finite number of directions to consider. You would then want to be extra careful in your analysis, either leveraging more topology or cleverly arguing with the reverse triangle inequality.
$endgroup$
– GenericMathematician
Dec 9 '18 at 20:01
|
show 1 more comment
$begingroup$
I know very basic topology. When you have the interval $(-infty, p_0-r)$, is that because $p_0-r>p$, do we just assume that $p> -infty$? Also with the cases that you mentioned, do you mean when $q$ is in the ball surrounding $p$ and contained in the "line" connecting $p$ to $p_0$ and when $q$ is not on the "line" connecting $p_0$ to $p$?
$endgroup$
– K.M
Dec 9 '18 at 19:34
1
$begingroup$
@K.M We don't have to assume that $p>-infty$; $p$ is a real number, so it has to be larger than $-infty$. The cases I am referring to are when $p>p_0$ and $p<p_0$. This will allow you to get rid of those pesky absolute value bars.
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:37
$begingroup$
I thought that $p$ and $p_0$ are points, do $p$ and $p_0$ denote the distances from $0$, respectively?
$endgroup$
– K.M
Dec 9 '18 at 19:43
$begingroup$
@K.M They are both specific points, but they are points in $mathbb{R}$, so we can compare them, add them, subtract them, etc. There is no difference between a point and a signed distance from $0$ in $mathbb{R}$.
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:46
1
$begingroup$
@K.M If we were working in $mathbb{R}^n$ then "absolute values" really mean euclidean distances, and then you can't break into cases because there is no finite number of directions to consider. You would then want to be extra careful in your analysis, either leveraging more topology or cleverly arguing with the reverse triangle inequality.
$endgroup$
– GenericMathematician
Dec 9 '18 at 20:01
$begingroup$
I know very basic topology. When you have the interval $(-infty, p_0-r)$, is that because $p_0-r>p$, do we just assume that $p> -infty$? Also with the cases that you mentioned, do you mean when $q$ is in the ball surrounding $p$ and contained in the "line" connecting $p$ to $p_0$ and when $q$ is not on the "line" connecting $p_0$ to $p$?
$endgroup$
– K.M
Dec 9 '18 at 19:34
$begingroup$
I know very basic topology. When you have the interval $(-infty, p_0-r)$, is that because $p_0-r>p$, do we just assume that $p> -infty$? Also with the cases that you mentioned, do you mean when $q$ is in the ball surrounding $p$ and contained in the "line" connecting $p$ to $p_0$ and when $q$ is not on the "line" connecting $p_0$ to $p$?
$endgroup$
– K.M
Dec 9 '18 at 19:34
1
1
$begingroup$
@K.M We don't have to assume that $p>-infty$; $p$ is a real number, so it has to be larger than $-infty$. The cases I am referring to are when $p>p_0$ and $p<p_0$. This will allow you to get rid of those pesky absolute value bars.
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:37
$begingroup$
@K.M We don't have to assume that $p>-infty$; $p$ is a real number, so it has to be larger than $-infty$. The cases I am referring to are when $p>p_0$ and $p<p_0$. This will allow you to get rid of those pesky absolute value bars.
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:37
$begingroup$
I thought that $p$ and $p_0$ are points, do $p$ and $p_0$ denote the distances from $0$, respectively?
$endgroup$
– K.M
Dec 9 '18 at 19:43
$begingroup$
I thought that $p$ and $p_0$ are points, do $p$ and $p_0$ denote the distances from $0$, respectively?
$endgroup$
– K.M
Dec 9 '18 at 19:43
$begingroup$
@K.M They are both specific points, but they are points in $mathbb{R}$, so we can compare them, add them, subtract them, etc. There is no difference between a point and a signed distance from $0$ in $mathbb{R}$.
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:46
$begingroup$
@K.M They are both specific points, but they are points in $mathbb{R}$, so we can compare them, add them, subtract them, etc. There is no difference between a point and a signed distance from $0$ in $mathbb{R}$.
$endgroup$
– GenericMathematician
Dec 9 '18 at 19:46
1
1
$begingroup$
@K.M If we were working in $mathbb{R}^n$ then "absolute values" really mean euclidean distances, and then you can't break into cases because there is no finite number of directions to consider. You would then want to be extra careful in your analysis, either leveraging more topology or cleverly arguing with the reverse triangle inequality.
$endgroup$
– GenericMathematician
Dec 9 '18 at 20:01
$begingroup$
@K.M If we were working in $mathbb{R}^n$ then "absolute values" really mean euclidean distances, and then you can't break into cases because there is no finite number of directions to consider. You would then want to be extra careful in your analysis, either leveraging more topology or cleverly arguing with the reverse triangle inequality.
$endgroup$
– GenericMathematician
Dec 9 '18 at 20:01
|
show 1 more comment
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