[Proof Verification]: The closure of a set is closed.
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Definition: The closure of a set $A$ is $bar A=Acup A'$, where $A'$
is the set of all limit points of $A$.
Claim: $bar A$ is a closed set.
Proof: (my attempt) If $bar A$ is a closed set then that implies that it contains all its limit points. So suppose to the contrary that $bar A$ is not a closed set. Then $exists$ a limit point $p$ of $bar A$ such that $pnot in bar A.$ Clearly, $p$ is not a limit point of $A$ because if it were then $pin bar A$. This means that $exists$ a neighborhood $N_r(p)$ which does not contain any point of $A$. But $p$ is a limit point of $bar A$ so it must contain an element $yin bar A-A$ in its neighborhood $N_r(p).$ Of course, $y$ is a limit point. Now, $0<d(p,y)=h<r.$ If we choose $0<epsilon<r-h$, then $N_epsilon(y)$ will contain no point $xin A$, which is a contradiction since $y$ is a limit point.
I want to know if this proof is correct or not?
real-analysis proof-verification
edited Jan 23 '17 at 15:47
Stella Biderman
26.2k63175
asked Jan 23 '17 at 15:35
Hello_World
3,37821429
add a comment |Â
up vote
4
down vote
favorite
Definition: The closure of a set $A$ is $bar A=Acup A'$, where $A'$
is the set of all limit points of $A$.
Claim: $bar A$ is a closed set.
Proof: (my attempt) If $bar A$ is a closed set then that implies that it contains all its limit points. So suppose to the contrary that $bar A$ is not a closed set. Then $exists$ a limit point $p$ of $bar A$ such that $pnot in bar A.$ Clearly, $p$ is not a limit point of $A$ because if it were then $pin bar A$. This means that $exists$ a neighborhood $N_r(p)$ which does not contain any point of $A$. But $p$ is a limit point of $bar A$ so it must contain an element $yin bar A-A$ in its neighborhood $N_r(p).$ Of course, $y$ is a limit point. Now, $0<d(p,y)=h<r.$ If we choose $0<epsilon<r-h$, then $N_epsilon(y)$ will contain no point $xin A$, which is a contradiction since $y$ is a limit point.
I want to know if this proof is correct or not?
real-analysis proof-verification
edited Jan 23 '17 at 15:47
Stella Biderman
26.2k63175
asked Jan 23 '17 at 15:35
Hello_World
3,37821429
yes, the proof is very good.
– Jorge Fernández
Jan 23 '17 at 15:37
Look at this for additional reference math.stackexchange.com/questions/448468/…
– Î˜Î£Î¦GenSan
Jan 23 '17 at 15:37
1
You may define the closure of $A$ as the smallest (with respect to $subseteq$) closed set enclosing $A$. In such a way the claim is trivial.
– Jack D'Aurizio♦
Jan 23 '17 at 15:39
@JackD'Aurizio But then we have to prove that your closure is really $Acup A'$. The point of this exercise is not really to show that the closure is closed (however it happens to be defined), but rather that the operation $Amapsto Acup A'$ (no matter what name it has) gives a closed set regardless of $A$.
– Arthur
Jan 23 '17 at 16:21
You do not need any use of a metric.This is true in any topological space, whether metrizable or not.
– DanielWainfleet
Apr 15 at 12:22
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Definition: The closure of a set $A$ is $bar A=Acup A'$, where $A'$
is the set of all limit points of $A$.
Claim: $bar A$ is a closed set.
Proof: (my attempt) If $bar A$ is a closed set then that implies that it contains all its limit points. So suppose to the contrary that $bar A$ is not a closed set. Then $exists$ a limit point $p$ of $bar A$ such that $pnot in bar A.$ Clearly, $p$ is not a limit point of $A$ because if it were then $pin bar A$. This means that $exists$ a neighborhood $N_r(p)$ which does not contain any point of $A$. But $p$ is a limit point of $bar A$ so it must contain an element $yin bar A-A$ in its neighborhood $N_r(p).$ Of course, $y$ is a limit point. Now, $0<d(p,y)=h<r.$ If we choose $0<epsilon<r-h$, then $N_epsilon(y)$ will contain no point $xin A$, which is a contradiction since $y$ is a limit point.
I want to know if this proof is correct or not?
real-analysis proof-verification
edited Jan 23 '17 at 15:47
Stella Biderman
26.2k63175
asked Jan 23 '17 at 15:35
Hello_World
3,37821429
Definition: The closure of a set $A$ is $bar A=Acup A'$, where $A'$
is the set of all limit points of $A$.
Claim: $bar A$ is a closed set.
Proof: (my attempt) If $bar A$ is a closed set then that implies that it contains all its limit points. So suppose to the contrary that $bar A$ is not a closed set. Then $exists$ a limit point $p$ of $bar A$ such that $pnot in bar A.$ Clearly, $p$ is not a limit point of $A$ because if it were then $pin bar A$. This means that $exists$ a neighborhood $N_r(p)$ which does not contain any point of $A$. But $p$ is a limit point of $bar A$ so it must contain an element $yin bar A-A$ in its neighborhood $N_r(p).$ Of course, $y$ is a limit point. Now, $0<d(p,y)=h<r.$ If we choose $0<epsilon<r-h$, then $N_epsilon(y)$ will contain no point $xin A$, which is a contradiction since $y$ is a limit point.
I want to know if this proof is correct or not?
real-analysis proof-verification
real-analysis proof-verification
edited Jan 23 '17 at 15:47
Stella Biderman
26.2k63175
asked Jan 23 '17 at 15:35
Hello_World
3,37821429
edited Jan 23 '17 at 15:47
Stella Biderman
26.2k63175
asked Jan 23 '17 at 15:35
Hello_World
3,37821429
edited Jan 23 '17 at 15:47
Stella Biderman
26.2k63175
edited Jan 23 '17 at 15:47
Stella Biderman
26.2k63175
edited Jan 23 '17 at 15:47
Stella Biderman
26.2k63175
26.2k63175
asked Jan 23 '17 at 15:35
Hello_World
3,37821429
asked Jan 23 '17 at 15:35
Hello_World
3,37821429
asked Jan 23 '17 at 15:35
Hello_World
3,37821429
3,37821429
yes, the proof is very good.
– Jorge Fernández
Jan 23 '17 at 15:37
Look at this for additional reference math.stackexchange.com/questions/448468/…
– Î˜Î£Î¦GenSan
Jan 23 '17 at 15:37
1
You may define the closure of $A$ as the smallest (with respect to $subseteq$) closed set enclosing $A$. In such a way the claim is trivial.
– Jack D'Aurizio♦
Jan 23 '17 at 15:39
@JackD'Aurizio But then we have to prove that your closure is really $Acup A'$. The point of this exercise is not really to show that the closure is closed (however it happens to be defined), but rather that the operation $Amapsto Acup A'$ (no matter what name it has) gives a closed set regardless of $A$.
– Arthur
Jan 23 '17 at 16:21
You do not need any use of a metric.This is true in any topological space, whether metrizable or not.
– DanielWainfleet
Apr 15 at 12:22
add a comment |Â
yes, the proof is very good.
– Jorge Fernández
Jan 23 '17 at 15:37
Look at this for additional reference math.stackexchange.com/questions/448468/…
– Î˜Î£Î¦GenSan
Jan 23 '17 at 15:37
1
You may define the closure of $A$ as the smallest (with respect to $subseteq$) closed set enclosing $A$. In such a way the claim is trivial.
– Jack D'Aurizio♦
Jan 23 '17 at 15:39
@JackD'Aurizio But then we have to prove that your closure is really $Acup A'$. The point of this exercise is not really to show that the closure is closed (however it happens to be defined), but rather that the operation $Amapsto Acup A'$ (no matter what name it has) gives a closed set regardless of $A$.
– Arthur
Jan 23 '17 at 16:21
You do not need any use of a metric.This is true in any topological space, whether metrizable or not.
– DanielWainfleet
Apr 15 at 12:22
yes, the proof is very good.
– Jorge Fernández
Jan 23 '17 at 15:37
yes, the proof is very good.
– Jorge Fernández
Jan 23 '17 at 15:37
Look at this for additional reference math.stackexchange.com/questions/448468/…
– Î˜Î£Î¦GenSan
Jan 23 '17 at 15:37
Look at this for additional reference math.stackexchange.com/questions/448468/…
– Î˜Î£Î¦GenSan
Jan 23 '17 at 15:37
1
1
You may define the closure of $A$ as the smallest (with respect to $subseteq$) closed set enclosing $A$. In such a way the claim is trivial.
– Jack D'Aurizio♦
Jan 23 '17 at 15:39
You may define the closure of $A$ as the smallest (with respect to $subseteq$) closed set enclosing $A$. In such a way the claim is trivial.
– Jack D'Aurizio♦
Jan 23 '17 at 15:39
@JackD'Aurizio But then we have to prove that your closure is really $Acup A'$. The point of this exercise is not really to show that the closure is closed (however it happens to be defined), but rather that the operation $Amapsto Acup A'$ (no matter what name it has) gives a closed set regardless of $A$.
– Arthur
Jan 23 '17 at 16:21
@JackD'Aurizio But then we have to prove that your closure is really $Acup A'$. The point of this exercise is not really to show that the closure is closed (however it happens to be defined), but rather that the operation $Amapsto Acup A'$ (no matter what name it has) gives a closed set regardless of $A$.
– Arthur
Jan 23 '17 at 16:21
You do not need any use of a metric.This is true in any topological space, whether metrizable or not.
– DanielWainfleet
Apr 15 at 12:22
You do not need any use of a metric.This is true in any topological space, whether metrizable or not.
– DanielWainfleet
Apr 15 at 12:22
add a comment |Â
2 Answers
2
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oldest
votes
up vote
0
down vote
Your proof is correct, maybe we can make it slightly faster.
Let $z$ be a limit point of $overline A$. Every open set $U$ must contain a point $x$ in $overline A$, if the point is in $A^circ$ then $U$ must intersect $A$ because it contains a limit point of $A$. if $x$ is in $A$ then $U$ also intersects $A$.
So every limit point of $overline A$ is a limit point of $A$, and $overline A$ contains all of its limit points.
answered Jan 23 '17 at 15:41
Jorge Fernández
74.3k1088186
add a comment |Â
up vote
0
down vote
If z is a limit point of $barA$ and U is a open set containing z then it must contain a point in $barA$. But as an open set containing a point in $barA$ it must contain a point in A.
answered Sep 6 at 11:34
StephaneM
211
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Your proof is correct, maybe we can make it slightly faster.
Let $z$ be a limit point of $overline A$. Every open set $U$ must contain a point $x$ in $overline A$, if the point is in $A^circ$ then $U$ must intersect $A$ because it contains a limit point of $A$. if $x$ is in $A$ then $U$ also intersects $A$.
So every limit point of $overline A$ is a limit point of $A$, and $overline A$ contains all of its limit points.
answered Jan 23 '17 at 15:41
Jorge Fernández
74.3k1088186
add a comment |Â
up vote
0
down vote
Your proof is correct, maybe we can make it slightly faster.
Let $z$ be a limit point of $overline A$. Every open set $U$ must contain a point $x$ in $overline A$, if the point is in $A^circ$ then $U$ must intersect $A$ because it contains a limit point of $A$. if $x$ is in $A$ then $U$ also intersects $A$.
So every limit point of $overline A$ is a limit point of $A$, and $overline A$ contains all of its limit points.
answered Jan 23 '17 at 15:41
Jorge Fernández
74.3k1088186
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your proof is correct, maybe we can make it slightly faster.
Let $z$ be a limit point of $overline A$. Every open set $U$ must contain a point $x$ in $overline A$, if the point is in $A^circ$ then $U$ must intersect $A$ because it contains a limit point of $A$. if $x$ is in $A$ then $U$ also intersects $A$.
So every limit point of $overline A$ is a limit point of $A$, and $overline A$ contains all of its limit points.
answered Jan 23 '17 at 15:41
Jorge Fernández
74.3k1088186
Your proof is correct, maybe we can make it slightly faster.
Let $z$ be a limit point of $overline A$. Every open set $U$ must contain a point $x$ in $overline A$, if the point is in $A^circ$ then $U$ must intersect $A$ because it contains a limit point of $A$. if $x$ is in $A$ then $U$ also intersects $A$.
So every limit point of $overline A$ is a limit point of $A$, and $overline A$ contains all of its limit points.
answered Jan 23 '17 at 15:41
Jorge Fernández
74.3k1088186
answered Jan 23 '17 at 15:41
Jorge Fernández
74.3k1088186
answered Jan 23 '17 at 15:41
Jorge Fernández
74.3k1088186
answered Jan 23 '17 at 15:41
Jorge Fernández
74.3k1088186
74.3k1088186
add a comment |Â
add a comment |Â
up vote
0
down vote
If z is a limit point of $barA$ and U is a open set containing z then it must contain a point in $barA$. But as an open set containing a point in $barA$ it must contain a point in A.
answered Sep 6 at 11:34
StephaneM
211
add a comment |Â
up vote
0
down vote
If z is a limit point of $barA$ and U is a open set containing z then it must contain a point in $barA$. But as an open set containing a point in $barA$ it must contain a point in A.
answered Sep 6 at 11:34
StephaneM
211
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If z is a limit point of $barA$ and U is a open set containing z then it must contain a point in $barA$. But as an open set containing a point in $barA$ it must contain a point in A.
answered Sep 6 at 11:34
StephaneM
211
If z is a limit point of $barA$ and U is a open set containing z then it must contain a point in $barA$. But as an open set containing a point in $barA$ it must contain a point in A.
answered Sep 6 at 11:34
StephaneM
211
answered Sep 6 at 11:34
StephaneM
211
answered Sep 6 at 11:34
StephaneM
211
answered Sep 6 at 11:34
StephaneM
211
211
add a comment |Â
add a comment |Â
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yes, the proof is very good.
– Jorge Fernández
Jan 23 '17 at 15:37
Look at this for additional reference math.stackexchange.com/questions/448468/…
– Î˜Î£Î¦GenSan
Jan 23 '17 at 15:37
1
You may define the closure of $A$ as the smallest (with respect to $subseteq$) closed set enclosing $A$. In such a way the claim is trivial.
– Jack D'Aurizio♦
Jan 23 '17 at 15:39
@JackD'Aurizio But then we have to prove that your closure is really $Acup A'$. The point of this exercise is not really to show that the closure is closed (however it happens to be defined), but rather that the operation $Amapsto Acup A'$ (no matter what name it has) gives a closed set regardless of $A$.
– Arthur
Jan 23 '17 at 16:21
You do not need any use of a metric.This is true in any topological space, whether metrizable or not.
– DanielWainfleet
Apr 15 at 12:22