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Real Analysis - Limit points and Open set.
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The statement is If $A subset mathbbR $ is any set, $I$ is an open set that contains a limit point of $L_A$ - the set of all limit points of $A$, then $I$ also contains a point of $A$. I approach this by letting $x$ in A with $x$ is a limit point of A. So for every $epsilon >0$ the deleted neighborhood $V^*(a) cap A$ is nonempty. From this how do I show that $x$ is also in $I$.
real-analysis general-topology
asked 58 mins ago
Dong Le
365
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up vote
2
down vote
favorite
The statement is If $A subset mathbbR $ is any set, $I$ is an open set that contains a limit point of $L_A$ - the set of all limit points of $A$, then $I$ also contains a point of $A$. I approach this by letting $x$ in A with $x$ is a limit point of A. So for every $epsilon >0$ the deleted neighborhood $V^*(a) cap A$ is nonempty. From this how do I show that $x$ is also in $I$.
real-analysis general-topology
asked 58 mins ago
Dong Le
365
it is a point of $L_A$.
– Dong Le
47 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The statement is If $A subset mathbbR $ is any set, $I$ is an open set that contains a limit point of $L_A$ - the set of all limit points of $A$, then $I$ also contains a point of $A$. I approach this by letting $x$ in A with $x$ is a limit point of A. So for every $epsilon >0$ the deleted neighborhood $V^*(a) cap A$ is nonempty. From this how do I show that $x$ is also in $I$.
real-analysis general-topology
asked 58 mins ago
Dong Le
365
The statement is If $A subset mathbbR $ is any set, $I$ is an open set that contains a limit point of $L_A$ - the set of all limit points of $A$, then $I$ also contains a point of $A$. I approach this by letting $x$ in A with $x$ is a limit point of A. So for every $epsilon >0$ the deleted neighborhood $V^*(a) cap A$ is nonempty. From this how do I show that $x$ is also in $I$.
real-analysis general-topology
real-analysis general-topology
asked 58 mins ago
Dong Le
365
asked 58 mins ago
Dong Le
365
asked 58 mins ago
Dong Le
365
asked 58 mins ago
Dong Le
365
asked 58 mins ago
Dong Le
365
365
it is a point of $L_A$.
– Dong Le
47 mins ago
add a comment |Â
it is a point of $L_A$.
– Dong Le
47 mins ago
it is a point of $L_A$.
– Dong Le
47 mins ago
it is a point of $L_A$.
– Dong Le
47 mins ago
add a comment |Â
2 Answers
2
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oldest
votes
up vote
3
down vote
If $I$ is an open set containing a limit point of $A$, call it $x$ as you have. Then since $I$ is open it contains a ball around $x$, say $(x-epsilon,x+epsilon)$. Then since $x$ is a limit point of $A$ we know that as you've already pointed out $((x-epsilon,x+epsilon))cap Anot=emptyset$, so since $((x-epsilon,x+epsilon))subseteq I$ it follows that $((x-epsilon,x+epsilon))cap Asubseteq I$, hence $I$ contains an element of $A$.
answered 51 mins ago
Melody
1739
1
I forgot to mention that the deleted neighborhood is $V^*_epsilon(x) = (x-epsilon, x) cup (x, x+epsilon)$ which basically removes the element $x$ from the ball.
– Dong Le
48 mins ago
Yes, I realized my error. It's been corrected to adjust for the fact that we don't know $xin A$, merely that it is a limit point.
– Melody
46 mins ago
1
I see! Thank you very much!!
– Dong Le
44 mins ago
Actually, I'm gonna change it back. It doesn't matter if we consider the deleted neighborhood or not. All we need to know is the neighborhood of $x$ has a point in $A$, as the entire neighborhood is in $I$.
– Melody
42 mins ago
add a comment |Â
up vote
1
down vote
Let $X$ be any topological space and $Asubset X.$ As per the comments by the proposer, a point $xin X$ is a limit point of $A$ iff every nbhd of $x$ contains a member of $A.$
Suppose $y$ is a limit point of the set $L_A$ of limit points of $A$. If $U$ is any nbhd of $y$ then there is an open $V$ with $yin Vsubset U.$
Now $V$ is a nbhd of $y$ so there exists $xin Vcap L_A.$ But $V$ is also a nbhd of $x$ (because $V$ is open and $xin V$ ), and $xin L_A$, so there exists $ain Acap V.$
So $ain Acap U$ because $Vsubset U.$
So any nbhd $U$ of $y$ contains a member of $A.$ So $yin L_A.$
This holds in every topological space.
answered 24 mins ago
DanielWainfleet
32.9k31645
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
If $I$ is an open set containing a limit point of $A$, call it $x$ as you have. Then since $I$ is open it contains a ball around $x$, say $(x-epsilon,x+epsilon)$. Then since $x$ is a limit point of $A$ we know that as you've already pointed out $((x-epsilon,x+epsilon))cap Anot=emptyset$, so since $((x-epsilon,x+epsilon))subseteq I$ it follows that $((x-epsilon,x+epsilon))cap Asubseteq I$, hence $I$ contains an element of $A$.
answered 51 mins ago
Melody
1739
1
I forgot to mention that the deleted neighborhood is $V^*_epsilon(x) = (x-epsilon, x) cup (x, x+epsilon)$ which basically removes the element $x$ from the ball.
– Dong Le
48 mins ago
Yes, I realized my error. It's been corrected to adjust for the fact that we don't know $xin A$, merely that it is a limit point.
– Melody
46 mins ago
1
I see! Thank you very much!!
– Dong Le
44 mins ago
Actually, I'm gonna change it back. It doesn't matter if we consider the deleted neighborhood or not. All we need to know is the neighborhood of $x$ has a point in $A$, as the entire neighborhood is in $I$.
– Melody
42 mins ago
add a comment |Â
up vote
3
down vote
If $I$ is an open set containing a limit point of $A$, call it $x$ as you have. Then since $I$ is open it contains a ball around $x$, say $(x-epsilon,x+epsilon)$. Then since $x$ is a limit point of $A$ we know that as you've already pointed out $((x-epsilon,x+epsilon))cap Anot=emptyset$, so since $((x-epsilon,x+epsilon))subseteq I$ it follows that $((x-epsilon,x+epsilon))cap Asubseteq I$, hence $I$ contains an element of $A$.
answered 51 mins ago
Melody
1739
1
I forgot to mention that the deleted neighborhood is $V^*_epsilon(x) = (x-epsilon, x) cup (x, x+epsilon)$ which basically removes the element $x$ from the ball.
– Dong Le
48 mins ago
Yes, I realized my error. It's been corrected to adjust for the fact that we don't know $xin A$, merely that it is a limit point.
– Melody
46 mins ago
1
I see! Thank you very much!!
– Dong Le
44 mins ago
Actually, I'm gonna change it back. It doesn't matter if we consider the deleted neighborhood or not. All we need to know is the neighborhood of $x$ has a point in $A$, as the entire neighborhood is in $I$.
– Melody
42 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If $I$ is an open set containing a limit point of $A$, call it $x$ as you have. Then since $I$ is open it contains a ball around $x$, say $(x-epsilon,x+epsilon)$. Then since $x$ is a limit point of $A$ we know that as you've already pointed out $((x-epsilon,x+epsilon))cap Anot=emptyset$, so since $((x-epsilon,x+epsilon))subseteq I$ it follows that $((x-epsilon,x+epsilon))cap Asubseteq I$, hence $I$ contains an element of $A$.
answered 51 mins ago
Melody
1739
If $I$ is an open set containing a limit point of $A$, call it $x$ as you have. Then since $I$ is open it contains a ball around $x$, say $(x-epsilon,x+epsilon)$. Then since $x$ is a limit point of $A$ we know that as you've already pointed out $((x-epsilon,x+epsilon))cap Anot=emptyset$, so since $((x-epsilon,x+epsilon))subseteq I$ it follows that $((x-epsilon,x+epsilon))cap Asubseteq I$, hence $I$ contains an element of $A$.
answered 51 mins ago
Melody
1739
edited 41 mins ago
answered 51 mins ago
Melody
1739
answered 51 mins ago
Melody
1739
answered 51 mins ago
Melody
1739
1739
1
I forgot to mention that the deleted neighborhood is $V^*_epsilon(x) = (x-epsilon, x) cup (x, x+epsilon)$ which basically removes the element $x$ from the ball.
– Dong Le
48 mins ago
Yes, I realized my error. It's been corrected to adjust for the fact that we don't know $xin A$, merely that it is a limit point.
– Melody
46 mins ago
1
I see! Thank you very much!!
– Dong Le
44 mins ago
Actually, I'm gonna change it back. It doesn't matter if we consider the deleted neighborhood or not. All we need to know is the neighborhood of $x$ has a point in $A$, as the entire neighborhood is in $I$.
– Melody
42 mins ago
add a comment |Â
1
I forgot to mention that the deleted neighborhood is $V^*_epsilon(x) = (x-epsilon, x) cup (x, x+epsilon)$ which basically removes the element $x$ from the ball.
– Dong Le
48 mins ago
Yes, I realized my error. It's been corrected to adjust for the fact that we don't know $xin A$, merely that it is a limit point.
– Melody
46 mins ago
1
I see! Thank you very much!!
– Dong Le
44 mins ago
Actually, I'm gonna change it back. It doesn't matter if we consider the deleted neighborhood or not. All we need to know is the neighborhood of $x$ has a point in $A$, as the entire neighborhood is in $I$.
– Melody
42 mins ago
1
1
I forgot to mention that the deleted neighborhood is $V^*_epsilon(x) = (x-epsilon, x) cup (x, x+epsilon)$ which basically removes the element $x$ from the ball.
– Dong Le
48 mins ago
I forgot to mention that the deleted neighborhood is $V^*_epsilon(x) = (x-epsilon, x) cup (x, x+epsilon)$ which basically removes the element $x$ from the ball.
– Dong Le
48 mins ago
Yes, I realized my error. It's been corrected to adjust for the fact that we don't know $xin A$, merely that it is a limit point.
– Melody
46 mins ago
Yes, I realized my error. It's been corrected to adjust for the fact that we don't know $xin A$, merely that it is a limit point.
– Melody
46 mins ago
1
1
I see! Thank you very much!!
– Dong Le
44 mins ago
I see! Thank you very much!!
– Dong Le
44 mins ago
Actually, I'm gonna change it back. It doesn't matter if we consider the deleted neighborhood or not. All we need to know is the neighborhood of $x$ has a point in $A$, as the entire neighborhood is in $I$.
– Melody
42 mins ago
Actually, I'm gonna change it back. It doesn't matter if we consider the deleted neighborhood or not. All we need to know is the neighborhood of $x$ has a point in $A$, as the entire neighborhood is in $I$.
– Melody
42 mins ago
add a comment |Â
up vote
1
down vote
Let $X$ be any topological space and $Asubset X.$ As per the comments by the proposer, a point $xin X$ is a limit point of $A$ iff every nbhd of $x$ contains a member of $A.$
Suppose $y$ is a limit point of the set $L_A$ of limit points of $A$. If $U$ is any nbhd of $y$ then there is an open $V$ with $yin Vsubset U.$
Now $V$ is a nbhd of $y$ so there exists $xin Vcap L_A.$ But $V$ is also a nbhd of $x$ (because $V$ is open and $xin V$ ), and $xin L_A$, so there exists $ain Acap V.$
So $ain Acap U$ because $Vsubset U.$
So any nbhd $U$ of $y$ contains a member of $A.$ So $yin L_A.$
This holds in every topological space.
answered 24 mins ago
DanielWainfleet
32.9k31645
add a comment |Â
up vote
1
down vote
Let $X$ be any topological space and $Asubset X.$ As per the comments by the proposer, a point $xin X$ is a limit point of $A$ iff every nbhd of $x$ contains a member of $A.$
Suppose $y$ is a limit point of the set $L_A$ of limit points of $A$. If $U$ is any nbhd of $y$ then there is an open $V$ with $yin Vsubset U.$
Now $V$ is a nbhd of $y$ so there exists $xin Vcap L_A.$ But $V$ is also a nbhd of $x$ (because $V$ is open and $xin V$ ), and $xin L_A$, so there exists $ain Acap V.$
So $ain Acap U$ because $Vsubset U.$
So any nbhd $U$ of $y$ contains a member of $A.$ So $yin L_A.$
This holds in every topological space.
answered 24 mins ago
DanielWainfleet
32.9k31645
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $X$ be any topological space and $Asubset X.$ As per the comments by the proposer, a point $xin X$ is a limit point of $A$ iff every nbhd of $x$ contains a member of $A.$
Suppose $y$ is a limit point of the set $L_A$ of limit points of $A$. If $U$ is any nbhd of $y$ then there is an open $V$ with $yin Vsubset U.$
Now $V$ is a nbhd of $y$ so there exists $xin Vcap L_A.$ But $V$ is also a nbhd of $x$ (because $V$ is open and $xin V$ ), and $xin L_A$, so there exists $ain Acap V.$
So $ain Acap U$ because $Vsubset U.$
So any nbhd $U$ of $y$ contains a member of $A.$ So $yin L_A.$
This holds in every topological space.
answered 24 mins ago
DanielWainfleet
32.9k31645
Let $X$ be any topological space and $Asubset X.$ As per the comments by the proposer, a point $xin X$ is a limit point of $A$ iff every nbhd of $x$ contains a member of $A.$
Suppose $y$ is a limit point of the set $L_A$ of limit points of $A$. If $U$ is any nbhd of $y$ then there is an open $V$ with $yin Vsubset U.$
Now $V$ is a nbhd of $y$ so there exists $xin Vcap L_A.$ But $V$ is also a nbhd of $x$ (because $V$ is open and $xin V$ ), and $xin L_A$, so there exists $ain Acap V.$
So $ain Acap U$ because $Vsubset U.$
So any nbhd $U$ of $y$ contains a member of $A.$ So $yin L_A.$
This holds in every topological space.
answered 24 mins ago
DanielWainfleet
32.9k31645
answered 24 mins ago
DanielWainfleet
32.9k31645
answered 24 mins ago
DanielWainfleet
32.9k31645
answered 24 mins ago
DanielWainfleet
32.9k31645
32.9k31645
add a comment |Â
add a comment |Â
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it is a point of $L_A$.
– Dong Le
47 mins ago