Proof Verification: Let $A$ be bounded above so that $s=$ sup $A$ exists. Prove that $s in overline{A}$.
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I have seen that this question has been answered already, but I wanted to come up with my own proof and would appreciate some feedback, as I am not always the best at transferring what is in my head into words and math symbology.
Proof: By definition, $overline{A} = A cup L$, where $L$ is the set of all limit points of $A$. In order for $s in overline{A}$, either $s in A$ or $s in L$.
Suppose $s in A$. Then we are done, as $s in A Rightarrow s in A cup L Rightarrow s in overline{A}$.
Now suppose that $s notin A$. By definition, as $s=$ sup $A$, $s$ is the least upper bound of $A$. Let $V_varepsilon(s)$ be an $varepsilon$-neighborhood of $s$. Then no matter how small we make $varepsilon$, $V_varepsilon(s) cap A neq emptyset$ since $s=$ sup $A$, which implies that $s in L$. Then $s in L Rightarrow sin A cup L Rightarrow sin overline{A}$.
Q.E.D.
Even if this proof is correct, I still don't feel great about the line where I write "Then, no matter how small we make $varepsilon$..." I guess it just doesn't feel very mathy to me, and I am also not sure it transitions as well as it could from the idea that $s=$ sup $A$.
real-analysis proof-verification
asked Nov 16 at 23:50
automattik
265
add a comment |
up vote
0
down vote
favorite
I have seen that this question has been answered already, but I wanted to come up with my own proof and would appreciate some feedback, as I am not always the best at transferring what is in my head into words and math symbology.
Proof: By definition, $overline{A} = A cup L$, where $L$ is the set of all limit points of $A$. In order for $s in overline{A}$, either $s in A$ or $s in L$.
Suppose $s in A$. Then we are done, as $s in A Rightarrow s in A cup L Rightarrow s in overline{A}$.
Now suppose that $s notin A$. By definition, as $s=$ sup $A$, $s$ is the least upper bound of $A$. Let $V_varepsilon(s)$ be an $varepsilon$-neighborhood of $s$. Then no matter how small we make $varepsilon$, $V_varepsilon(s) cap A neq emptyset$ since $s=$ sup $A$, which implies that $s in L$. Then $s in L Rightarrow sin A cup L Rightarrow sin overline{A}$.
Q.E.D.
Even if this proof is correct, I still don't feel great about the line where I write "Then, no matter how small we make $varepsilon$..." I guess it just doesn't feel very mathy to me, and I am also not sure it transitions as well as it could from the idea that $s=$ sup $A$.
real-analysis proof-verification
asked Nov 16 at 23:50
automattik
265
1
Your proof is quite 'mathy'!
– Kavi Rama Murthy
Nov 16 at 23:56
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have seen that this question has been answered already, but I wanted to come up with my own proof and would appreciate some feedback, as I am not always the best at transferring what is in my head into words and math symbology.
Proof: By definition, $overline{A} = A cup L$, where $L$ is the set of all limit points of $A$. In order for $s in overline{A}$, either $s in A$ or $s in L$.
Suppose $s in A$. Then we are done, as $s in A Rightarrow s in A cup L Rightarrow s in overline{A}$.
Now suppose that $s notin A$. By definition, as $s=$ sup $A$, $s$ is the least upper bound of $A$. Let $V_varepsilon(s)$ be an $varepsilon$-neighborhood of $s$. Then no matter how small we make $varepsilon$, $V_varepsilon(s) cap A neq emptyset$ since $s=$ sup $A$, which implies that $s in L$. Then $s in L Rightarrow sin A cup L Rightarrow sin overline{A}$.
Q.E.D.
Even if this proof is correct, I still don't feel great about the line where I write "Then, no matter how small we make $varepsilon$..." I guess it just doesn't feel very mathy to me, and I am also not sure it transitions as well as it could from the idea that $s=$ sup $A$.
real-analysis proof-verification
asked Nov 16 at 23:50
automattik
265
I have seen that this question has been answered already, but I wanted to come up with my own proof and would appreciate some feedback, as I am not always the best at transferring what is in my head into words and math symbology.
Proof: By definition, $overline{A} = A cup L$, where $L$ is the set of all limit points of $A$. In order for $s in overline{A}$, either $s in A$ or $s in L$.
Suppose $s in A$. Then we are done, as $s in A Rightarrow s in A cup L Rightarrow s in overline{A}$.
Now suppose that $s notin A$. By definition, as $s=$ sup $A$, $s$ is the least upper bound of $A$. Let $V_varepsilon(s)$ be an $varepsilon$-neighborhood of $s$. Then no matter how small we make $varepsilon$, $V_varepsilon(s) cap A neq emptyset$ since $s=$ sup $A$, which implies that $s in L$. Then $s in L Rightarrow sin A cup L Rightarrow sin overline{A}$.
Q.E.D.
Even if this proof is correct, I still don't feel great about the line where I write "Then, no matter how small we make $varepsilon$..." I guess it just doesn't feel very mathy to me, and I am also not sure it transitions as well as it could from the idea that $s=$ sup $A$.
real-analysis proof-verification
real-analysis proof-verification
asked Nov 16 at 23:50
automattik
265
asked Nov 16 at 23:50
automattik
265
asked Nov 16 at 23:50
automattik
265
asked Nov 16 at 23:50
automattik
265
asked Nov 16 at 23:50
automattik
265
265
1
Your proof is quite 'mathy'!
– Kavi Rama Murthy
Nov 16 at 23:56
add a comment |
1
Your proof is quite 'mathy'!
– Kavi Rama Murthy
Nov 16 at 23:56
1
1
Your proof is quite 'mathy'!
– Kavi Rama Murthy
Nov 16 at 23:56
Your proof is quite 'mathy'!
– Kavi Rama Murthy
Nov 16 at 23:56
add a comment |
1 Answer
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up vote
0
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I think "no matter how small we make $epsilon$" is how I would describe it if in conversation, but I can definitely get the desire to make it more 'mathy', particulary if it's for a graded assignment.
If you want to 'math it up a bit', my suggestion is that you spend a couple lines proving why $V_epsilon(s)cap Aneqemptyset$ for all $epsilon > 0$, possibly with a contradiction proof. For instance, if there is an $epsilon >0$ such that $V_epsilon(s)cap A =emptyset$, what can you say about some of the elements in this neighborhood and their relation to both $s$ and to all of the elements of $A$?
answered Nov 17 at 0:39
Tartaglia's Stutter
1495
I see what you mean... so maybe something along the lines of "If $V_varepsilon(s) cap A = emptyset$, this would imply that $s neq$ sup $A$, as $V_varepsilon(s)$ would contain elements less than $s$ that are not in $A$"?
– automattik
Nov 17 at 5:30
1
Not only not in $A$, but larger than every element in $A$. Which has consequences for the status of $s$ being $sup A$.
– Tartaglia's Stutter
Nov 17 at 11:53
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I think "no matter how small we make $epsilon$" is how I would describe it if in conversation, but I can definitely get the desire to make it more 'mathy', particulary if it's for a graded assignment.
If you want to 'math it up a bit', my suggestion is that you spend a couple lines proving why $V_epsilon(s)cap Aneqemptyset$ for all $epsilon > 0$, possibly with a contradiction proof. For instance, if there is an $epsilon >0$ such that $V_epsilon(s)cap A =emptyset$, what can you say about some of the elements in this neighborhood and their relation to both $s$ and to all of the elements of $A$?
answered Nov 17 at 0:39
Tartaglia's Stutter
1495
I see what you mean... so maybe something along the lines of "If $V_varepsilon(s) cap A = emptyset$, this would imply that $s neq$ sup $A$, as $V_varepsilon(s)$ would contain elements less than $s$ that are not in $A$"?
– automattik
Nov 17 at 5:30
1
Not only not in $A$, but larger than every element in $A$. Which has consequences for the status of $s$ being $sup A$.
– Tartaglia's Stutter
Nov 17 at 11:53
add a comment |
up vote
0
down vote
I think "no matter how small we make $epsilon$" is how I would describe it if in conversation, but I can definitely get the desire to make it more 'mathy', particulary if it's for a graded assignment.
If you want to 'math it up a bit', my suggestion is that you spend a couple lines proving why $V_epsilon(s)cap Aneqemptyset$ for all $epsilon > 0$, possibly with a contradiction proof. For instance, if there is an $epsilon >0$ such that $V_epsilon(s)cap A =emptyset$, what can you say about some of the elements in this neighborhood and their relation to both $s$ and to all of the elements of $A$?
answered Nov 17 at 0:39
Tartaglia's Stutter
1495
I see what you mean... so maybe something along the lines of "If $V_varepsilon(s) cap A = emptyset$, this would imply that $s neq$ sup $A$, as $V_varepsilon(s)$ would contain elements less than $s$ that are not in $A$"?
– automattik
Nov 17 at 5:30
1
Not only not in $A$, but larger than every element in $A$. Which has consequences for the status of $s$ being $sup A$.
– Tartaglia's Stutter
Nov 17 at 11:53
add a comment |
up vote
0
down vote
up vote
0
down vote
I think "no matter how small we make $epsilon$" is how I would describe it if in conversation, but I can definitely get the desire to make it more 'mathy', particulary if it's for a graded assignment.
If you want to 'math it up a bit', my suggestion is that you spend a couple lines proving why $V_epsilon(s)cap Aneqemptyset$ for all $epsilon > 0$, possibly with a contradiction proof. For instance, if there is an $epsilon >0$ such that $V_epsilon(s)cap A =emptyset$, what can you say about some of the elements in this neighborhood and their relation to both $s$ and to all of the elements of $A$?
answered Nov 17 at 0:39
Tartaglia's Stutter
1495
I think "no matter how small we make $epsilon$" is how I would describe it if in conversation, but I can definitely get the desire to make it more 'mathy', particulary if it's for a graded assignment.
If you want to 'math it up a bit', my suggestion is that you spend a couple lines proving why $V_epsilon(s)cap Aneqemptyset$ for all $epsilon > 0$, possibly with a contradiction proof. For instance, if there is an $epsilon >0$ such that $V_epsilon(s)cap A =emptyset$, what can you say about some of the elements in this neighborhood and their relation to both $s$ and to all of the elements of $A$?
answered Nov 17 at 0:39
Tartaglia's Stutter
1495
answered Nov 17 at 0:39
Tartaglia's Stutter
1495
answered Nov 17 at 0:39
Tartaglia's Stutter
1495
answered Nov 17 at 0:39
Tartaglia's Stutter
1495
1495
I see what you mean... so maybe something along the lines of "If $V_varepsilon(s) cap A = emptyset$, this would imply that $s neq$ sup $A$, as $V_varepsilon(s)$ would contain elements less than $s$ that are not in $A$"?
– automattik
Nov 17 at 5:30
1
Not only not in $A$, but larger than every element in $A$. Which has consequences for the status of $s$ being $sup A$.
– Tartaglia's Stutter
Nov 17 at 11:53
add a comment |
I see what you mean... so maybe something along the lines of "If $V_varepsilon(s) cap A = emptyset$, this would imply that $s neq$ sup $A$, as $V_varepsilon(s)$ would contain elements less than $s$ that are not in $A$"?
– automattik
Nov 17 at 5:30
1
Not only not in $A$, but larger than every element in $A$. Which has consequences for the status of $s$ being $sup A$.
– Tartaglia's Stutter
Nov 17 at 11:53
I see what you mean... so maybe something along the lines of "If $V_varepsilon(s) cap A = emptyset$, this would imply that $s neq$ sup $A$, as $V_varepsilon(s)$ would contain elements less than $s$ that are not in $A$"?
– automattik
Nov 17 at 5:30
I see what you mean... so maybe something along the lines of "If $V_varepsilon(s) cap A = emptyset$, this would imply that $s neq$ sup $A$, as $V_varepsilon(s)$ would contain elements less than $s$ that are not in $A$"?
– automattik
Nov 17 at 5:30
1
1
Not only not in $A$, but larger than every element in $A$. Which has consequences for the status of $s$ being $sup A$.
– Tartaglia's Stutter
Nov 17 at 11:53
Not only not in $A$, but larger than every element in $A$. Which has consequences for the status of $s$ being $sup A$.
– Tartaglia's Stutter
Nov 17 at 11:53
add a comment |
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Your proof is quite 'mathy'!
– Kavi Rama Murthy
Nov 16 at 23:56