Closure of the union = Union of closures
I have seen that $text{cl}(Acup B)=text{cl}(A)cup text{cl}(B)$. However I don't see why this is true. I can see the the right to left inclusion, but I can't see the inclusion from left to right. Say that I have an element $x$ contained only in two open sets one that intersects $Acup B$ only in $Asetminus B$ and and another that intersects only in $Bsetminus A$ isn't this a contradiction?
Edit: I have seen the proof but I still can't understand what is wrong with the counterexaple above
Thanks in advance
general-topology
edited Oct 27 '16 at 15:49
Hermès
1,809612
asked Oct 26 '16 at 15:55
TheGeometer
920519
add a comment |
I have seen that $text{cl}(Acup B)=text{cl}(A)cup text{cl}(B)$. However I don't see why this is true. I can see the the right to left inclusion, but I can't see the inclusion from left to right. Say that I have an element $x$ contained only in two open sets one that intersects $Acup B$ only in $Asetminus B$ and and another that intersects only in $Bsetminus A$ isn't this a contradiction?
Edit: I have seen the proof but I still can't understand what is wrong with the counterexaple above
Thanks in advance
general-topology
edited Oct 27 '16 at 15:49
Hermès
1,809612
asked Oct 26 '16 at 15:55
TheGeometer
920519
Choose some limit point of LHS and observe that it belong to the RHS.
– Masacroso
Oct 26 '16 at 16:06
add a comment |
I have seen that $text{cl}(Acup B)=text{cl}(A)cup text{cl}(B)$. However I don't see why this is true. I can see the the right to left inclusion, but I can't see the inclusion from left to right. Say that I have an element $x$ contained only in two open sets one that intersects $Acup B$ only in $Asetminus B$ and and another that intersects only in $Bsetminus A$ isn't this a contradiction?
Edit: I have seen the proof but I still can't understand what is wrong with the counterexaple above
Thanks in advance
general-topology
edited Oct 27 '16 at 15:49
Hermès
1,809612
asked Oct 26 '16 at 15:55
TheGeometer
920519
I have seen that $text{cl}(Acup B)=text{cl}(A)cup text{cl}(B)$. However I don't see why this is true. I can see the the right to left inclusion, but I can't see the inclusion from left to right. Say that I have an element $x$ contained only in two open sets one that intersects $Acup B$ only in $Asetminus B$ and and another that intersects only in $Bsetminus A$ isn't this a contradiction?
Edit: I have seen the proof but I still can't understand what is wrong with the counterexaple above
Thanks in advance
general-topology
general-topology
edited Oct 27 '16 at 15:49
Hermès
1,809612
asked Oct 26 '16 at 15:55
TheGeometer
920519
edited Oct 27 '16 at 15:49
Hermès
1,809612
asked Oct 26 '16 at 15:55
TheGeometer
920519
edited Oct 27 '16 at 15:49
Hermès
1,809612
edited Oct 27 '16 at 15:49
Hermès
1,809612
edited Oct 27 '16 at 15:49
Hermès
1,809612
1,809612
asked Oct 26 '16 at 15:55
TheGeometer
920519
asked Oct 26 '16 at 15:55
TheGeometer
920519
asked Oct 26 '16 at 15:55
TheGeometer
920519
920519
Choose some limit point of LHS and observe that it belong to the RHS.
– Masacroso
Oct 26 '16 at 16:06
add a comment |
Choose some limit point of LHS and observe that it belong to the RHS.
– Masacroso
Oct 26 '16 at 16:06
Choose some limit point of LHS and observe that it belong to the RHS.
– Masacroso
Oct 26 '16 at 16:06
Choose some limit point of LHS and observe that it belong to the RHS.
– Masacroso
Oct 26 '16 at 16:06
add a comment |
4 Answers
4
active
oldest
votes
(1) ($supset$) ::
begin{align*}
A subset A cup B implies text{cl}(A) subset text{cl}(A cup B)
\B subset A cup B implies text{cl}(B) subset text{cl}(A cup B)
end{align*}
therefore yielding that $text{cl}(A) cup text{cl}(B) subsettext{cl}(A cup B)$
(2) ($subset$) ::
The subset $text{cl}(A) cup text{cl}(B)$ is closed and both contains $A$ and $B$, therefore $A cup B subset text{cl}(A) cup text{cl}(B)$. $text{cl}(A cup B)$ is defined to be smallest closed set which contained $A cup B$, so that any closed set which contained $Acup B$ also contains $text{cl}(A cup B)$. Therefore $text{cl}(A cup B) subset text{cl}(A) cup text{cl}(B)$.
answered Oct 26 '16 at 16:12
Hermès
1,809612
Thanks for answer the proof makes sense but I still can't see what is wrong with my counterexample
– TheGeometer
Oct 26 '16 at 16:26
But how come $x$ in cl(A)? I have a neigbourhood of $x$ only intersecting at BA so this neigbourhood does not intersect A so it shouldn't be a limitpoint of A right?
– TheGeometer
Oct 26 '16 at 16:36
@TheGeometer Indeed, I've got confused! Define your two open sets to be $U_1$ and $U_2$. Then $U_1 cap U_2$ is a third open set containing $x$, that neither intersect $A$ or $B$. So $x notin text{cl}(A cup B)$ ;). So it's not a counterexample.
– Hermès
Oct 26 '16 at 16:44
Exactly! Thanks man I got so paranoid with that problem:P
– TheGeometer
Oct 26 '16 at 16:45
add a comment |
In your proposed counterexample, you've forgotten that open sets are closed under finite intersection.
So if $x$ has a neighborhood that only meets $Acup B$ in $Bsetminus A$ and a neighborhood that only meets $Acup B$ in $Asetminus B$, then the intersection of these neighborhoods doesn't meet $Acup B$ at all.
answered Oct 26 '16 at 16:43
Alex Kruckman
26.7k22556
add a comment |
Let $xin Cl(Acup B)$ then every open set containing $x$ intersects $Acup B$. Thus for $x in U_alpha$ where $U_alpha$ is open in $X$. If $U_alpha$ intersects $A$, then $x in Cl(A)$ else $x in Cl(B)$ either way $x in Cl(A)cup Cl(B)$.
answered Oct 26 '16 at 16:06
Riemann-bitcoin.
336111
This is because if $yin Cl(A)$ iff every open set containing $y$ intersects $A$.
– Riemann-bitcoin.
Oct 26 '16 at 16:10
I think I have a confusion with the definition. How I know it is that $x$ is a limit point of a subset $S$ of a topological space if every neigbourhood of $x$ intersects with $S$. So in our case if $x$ has a neigbourhood only intersecting $AB$ and another only intersecting $BA$ it should be a limit point only of $AcupB$ and not of A or B.Am I using wrond definition or saying something wrong?
– TheGeometer
Oct 26 '16 at 16:19
The $Cl(A)$ is the limit points unioned with the set A. So $Cl(A)=A cup A'$. I believe this is the confusion, though I am slightly confused on what your counterexample is saying.
– Riemann-bitcoin.
Oct 26 '16 at 16:57
add a comment |
Infinite point sequence from (A ∪ B) contains an infinite subsequence from A or contains an infinite subsequence from B
answered Oct 26 '16 at 16:00
kotomord
1,460626
ok but what is wrong wit my counterexample?
– TheGeometer
Oct 26 '16 at 16:04
add a comment |
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4 Answers
4
active
oldest
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4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
(1) ($supset$) ::
begin{align*}
A subset A cup B implies text{cl}(A) subset text{cl}(A cup B)
\B subset A cup B implies text{cl}(B) subset text{cl}(A cup B)
end{align*}
therefore yielding that $text{cl}(A) cup text{cl}(B) subsettext{cl}(A cup B)$
(2) ($subset$) ::
The subset $text{cl}(A) cup text{cl}(B)$ is closed and both contains $A$ and $B$, therefore $A cup B subset text{cl}(A) cup text{cl}(B)$. $text{cl}(A cup B)$ is defined to be smallest closed set which contained $A cup B$, so that any closed set which contained $Acup B$ also contains $text{cl}(A cup B)$. Therefore $text{cl}(A cup B) subset text{cl}(A) cup text{cl}(B)$.
answered Oct 26 '16 at 16:12
Hermès
1,809612
Thanks for answer the proof makes sense but I still can't see what is wrong with my counterexample
– TheGeometer
Oct 26 '16 at 16:26
But how come $x$ in cl(A)? I have a neigbourhood of $x$ only intersecting at BA so this neigbourhood does not intersect A so it shouldn't be a limitpoint of A right?
– TheGeometer
Oct 26 '16 at 16:36
@TheGeometer Indeed, I've got confused! Define your two open sets to be $U_1$ and $U_2$. Then $U_1 cap U_2$ is a third open set containing $x$, that neither intersect $A$ or $B$. So $x notin text{cl}(A cup B)$ ;). So it's not a counterexample.
– Hermès
Oct 26 '16 at 16:44
Exactly! Thanks man I got so paranoid with that problem:P
– TheGeometer
Oct 26 '16 at 16:45
add a comment |
(1) ($supset$) ::
begin{align*}
A subset A cup B implies text{cl}(A) subset text{cl}(A cup B)
\B subset A cup B implies text{cl}(B) subset text{cl}(A cup B)
end{align*}
therefore yielding that $text{cl}(A) cup text{cl}(B) subsettext{cl}(A cup B)$
(2) ($subset$) ::
The subset $text{cl}(A) cup text{cl}(B)$ is closed and both contains $A$ and $B$, therefore $A cup B subset text{cl}(A) cup text{cl}(B)$. $text{cl}(A cup B)$ is defined to be smallest closed set which contained $A cup B$, so that any closed set which contained $Acup B$ also contains $text{cl}(A cup B)$. Therefore $text{cl}(A cup B) subset text{cl}(A) cup text{cl}(B)$.
answered Oct 26 '16 at 16:12
Hermès
1,809612
Thanks for answer the proof makes sense but I still can't see what is wrong with my counterexample
– TheGeometer
Oct 26 '16 at 16:26
But how come $x$ in cl(A)? I have a neigbourhood of $x$ only intersecting at BA so this neigbourhood does not intersect A so it shouldn't be a limitpoint of A right?
– TheGeometer
Oct 26 '16 at 16:36
@TheGeometer Indeed, I've got confused! Define your two open sets to be $U_1$ and $U_2$. Then $U_1 cap U_2$ is a third open set containing $x$, that neither intersect $A$ or $B$. So $x notin text{cl}(A cup B)$ ;). So it's not a counterexample.
– Hermès
Oct 26 '16 at 16:44
Exactly! Thanks man I got so paranoid with that problem:P
– TheGeometer
Oct 26 '16 at 16:45
add a comment |
(1) ($supset$) ::
begin{align*}
A subset A cup B implies text{cl}(A) subset text{cl}(A cup B)
\B subset A cup B implies text{cl}(B) subset text{cl}(A cup B)
end{align*}
therefore yielding that $text{cl}(A) cup text{cl}(B) subsettext{cl}(A cup B)$
(2) ($subset$) ::
The subset $text{cl}(A) cup text{cl}(B)$ is closed and both contains $A$ and $B$, therefore $A cup B subset text{cl}(A) cup text{cl}(B)$. $text{cl}(A cup B)$ is defined to be smallest closed set which contained $A cup B$, so that any closed set which contained $Acup B$ also contains $text{cl}(A cup B)$. Therefore $text{cl}(A cup B) subset text{cl}(A) cup text{cl}(B)$.
answered Oct 26 '16 at 16:12
Hermès
1,809612
(1) ($supset$) ::
begin{align*}
A subset A cup B implies text{cl}(A) subset text{cl}(A cup B)
\B subset A cup B implies text{cl}(B) subset text{cl}(A cup B)
end{align*}
therefore yielding that $text{cl}(A) cup text{cl}(B) subsettext{cl}(A cup B)$
(2) ($subset$) ::
The subset $text{cl}(A) cup text{cl}(B)$ is closed and both contains $A$ and $B$, therefore $A cup B subset text{cl}(A) cup text{cl}(B)$. $text{cl}(A cup B)$ is defined to be smallest closed set which contained $A cup B$, so that any closed set which contained $Acup B$ also contains $text{cl}(A cup B)$. Therefore $text{cl}(A cup B) subset text{cl}(A) cup text{cl}(B)$.
answered Oct 26 '16 at 16:12
Hermès
1,809612
answered Oct 26 '16 at 16:12
Hermès
1,809612
answered Oct 26 '16 at 16:12
Hermès
1,809612
answered Oct 26 '16 at 16:12
Hermès
1,809612
1,809612
Thanks for answer the proof makes sense but I still can't see what is wrong with my counterexample
– TheGeometer
Oct 26 '16 at 16:26
But how come $x$ in cl(A)? I have a neigbourhood of $x$ only intersecting at BA so this neigbourhood does not intersect A so it shouldn't be a limitpoint of A right?
– TheGeometer
Oct 26 '16 at 16:36
@TheGeometer Indeed, I've got confused! Define your two open sets to be $U_1$ and $U_2$. Then $U_1 cap U_2$ is a third open set containing $x$, that neither intersect $A$ or $B$. So $x notin text{cl}(A cup B)$ ;). So it's not a counterexample.
– Hermès
Oct 26 '16 at 16:44
Exactly! Thanks man I got so paranoid with that problem:P
– TheGeometer
Oct 26 '16 at 16:45
add a comment |
Thanks for answer the proof makes sense but I still can't see what is wrong with my counterexample
– TheGeometer
Oct 26 '16 at 16:26
But how come $x$ in cl(A)? I have a neigbourhood of $x$ only intersecting at BA so this neigbourhood does not intersect A so it shouldn't be a limitpoint of A right?
– TheGeometer
Oct 26 '16 at 16:36
@TheGeometer Indeed, I've got confused! Define your two open sets to be $U_1$ and $U_2$. Then $U_1 cap U_2$ is a third open set containing $x$, that neither intersect $A$ or $B$. So $x notin text{cl}(A cup B)$ ;). So it's not a counterexample.
– Hermès
Oct 26 '16 at 16:44
Exactly! Thanks man I got so paranoid with that problem:P
– TheGeometer
Oct 26 '16 at 16:45
Thanks for answer the proof makes sense but I still can't see what is wrong with my counterexample
– TheGeometer
Oct 26 '16 at 16:26
Thanks for answer the proof makes sense but I still can't see what is wrong with my counterexample
– TheGeometer
Oct 26 '16 at 16:26
But how come $x$ in cl(A)? I have a neigbourhood of $x$ only intersecting at BA so this neigbourhood does not intersect A so it shouldn't be a limitpoint of A right?
– TheGeometer
Oct 26 '16 at 16:36
But how come $x$ in cl(A)? I have a neigbourhood of $x$ only intersecting at BA so this neigbourhood does not intersect A so it shouldn't be a limitpoint of A right?
– TheGeometer
Oct 26 '16 at 16:36
@TheGeometer Indeed, I've got confused! Define your two open sets to be $U_1$ and $U_2$. Then $U_1 cap U_2$ is a third open set containing $x$, that neither intersect $A$ or $B$. So $x notin text{cl}(A cup B)$ ;). So it's not a counterexample.
– Hermès
Oct 26 '16 at 16:44
@TheGeometer Indeed, I've got confused! Define your two open sets to be $U_1$ and $U_2$. Then $U_1 cap U_2$ is a third open set containing $x$, that neither intersect $A$ or $B$. So $x notin text{cl}(A cup B)$ ;). So it's not a counterexample.
– Hermès
Oct 26 '16 at 16:44
Exactly! Thanks man I got so paranoid with that problem:P
– TheGeometer
Oct 26 '16 at 16:45
Exactly! Thanks man I got so paranoid with that problem:P
– TheGeometer
Oct 26 '16 at 16:45
add a comment |
In your proposed counterexample, you've forgotten that open sets are closed under finite intersection.
So if $x$ has a neighborhood that only meets $Acup B$ in $Bsetminus A$ and a neighborhood that only meets $Acup B$ in $Asetminus B$, then the intersection of these neighborhoods doesn't meet $Acup B$ at all.
answered Oct 26 '16 at 16:43
Alex Kruckman
26.7k22556
add a comment |
In your proposed counterexample, you've forgotten that open sets are closed under finite intersection.
So if $x$ has a neighborhood that only meets $Acup B$ in $Bsetminus A$ and a neighborhood that only meets $Acup B$ in $Asetminus B$, then the intersection of these neighborhoods doesn't meet $Acup B$ at all.
answered Oct 26 '16 at 16:43
Alex Kruckman
26.7k22556
add a comment |
In your proposed counterexample, you've forgotten that open sets are closed under finite intersection.
So if $x$ has a neighborhood that only meets $Acup B$ in $Bsetminus A$ and a neighborhood that only meets $Acup B$ in $Asetminus B$, then the intersection of these neighborhoods doesn't meet $Acup B$ at all.
answered Oct 26 '16 at 16:43
Alex Kruckman
26.7k22556
In your proposed counterexample, you've forgotten that open sets are closed under finite intersection.
So if $x$ has a neighborhood that only meets $Acup B$ in $Bsetminus A$ and a neighborhood that only meets $Acup B$ in $Asetminus B$, then the intersection of these neighborhoods doesn't meet $Acup B$ at all.
answered Oct 26 '16 at 16:43
Alex Kruckman
26.7k22556
answered Oct 26 '16 at 16:43
Alex Kruckman
26.7k22556
answered Oct 26 '16 at 16:43
Alex Kruckman
26.7k22556
answered Oct 26 '16 at 16:43
Alex Kruckman
26.7k22556
26.7k22556
add a comment |
add a comment |
Let $xin Cl(Acup B)$ then every open set containing $x$ intersects $Acup B$. Thus for $x in U_alpha$ where $U_alpha$ is open in $X$. If $U_alpha$ intersects $A$, then $x in Cl(A)$ else $x in Cl(B)$ either way $x in Cl(A)cup Cl(B)$.
answered Oct 26 '16 at 16:06
Riemann-bitcoin.
336111
This is because if $yin Cl(A)$ iff every open set containing $y$ intersects $A$.
– Riemann-bitcoin.
Oct 26 '16 at 16:10
I think I have a confusion with the definition. How I know it is that $x$ is a limit point of a subset $S$ of a topological space if every neigbourhood of $x$ intersects with $S$. So in our case if $x$ has a neigbourhood only intersecting $AB$ and another only intersecting $BA$ it should be a limit point only of $AcupB$ and not of A or B.Am I using wrond definition or saying something wrong?
– TheGeometer
Oct 26 '16 at 16:19
The $Cl(A)$ is the limit points unioned with the set A. So $Cl(A)=A cup A'$. I believe this is the confusion, though I am slightly confused on what your counterexample is saying.
– Riemann-bitcoin.
Oct 26 '16 at 16:57
add a comment |
Let $xin Cl(Acup B)$ then every open set containing $x$ intersects $Acup B$. Thus for $x in U_alpha$ where $U_alpha$ is open in $X$. If $U_alpha$ intersects $A$, then $x in Cl(A)$ else $x in Cl(B)$ either way $x in Cl(A)cup Cl(B)$.
answered Oct 26 '16 at 16:06
Riemann-bitcoin.
336111
This is because if $yin Cl(A)$ iff every open set containing $y$ intersects $A$.
– Riemann-bitcoin.
Oct 26 '16 at 16:10
I think I have a confusion with the definition. How I know it is that $x$ is a limit point of a subset $S$ of a topological space if every neigbourhood of $x$ intersects with $S$. So in our case if $x$ has a neigbourhood only intersecting $AB$ and another only intersecting $BA$ it should be a limit point only of $AcupB$ and not of A or B.Am I using wrond definition or saying something wrong?
– TheGeometer
Oct 26 '16 at 16:19
The $Cl(A)$ is the limit points unioned with the set A. So $Cl(A)=A cup A'$. I believe this is the confusion, though I am slightly confused on what your counterexample is saying.
– Riemann-bitcoin.
Oct 26 '16 at 16:57
add a comment |
Let $xin Cl(Acup B)$ then every open set containing $x$ intersects $Acup B$. Thus for $x in U_alpha$ where $U_alpha$ is open in $X$. If $U_alpha$ intersects $A$, then $x in Cl(A)$ else $x in Cl(B)$ either way $x in Cl(A)cup Cl(B)$.
answered Oct 26 '16 at 16:06
Riemann-bitcoin.
336111
Let $xin Cl(Acup B)$ then every open set containing $x$ intersects $Acup B$. Thus for $x in U_alpha$ where $U_alpha$ is open in $X$. If $U_alpha$ intersects $A$, then $x in Cl(A)$ else $x in Cl(B)$ either way $x in Cl(A)cup Cl(B)$.
answered Oct 26 '16 at 16:06
Riemann-bitcoin.
336111
answered Oct 26 '16 at 16:06
Riemann-bitcoin.
336111
answered Oct 26 '16 at 16:06
Riemann-bitcoin.
336111
answered Oct 26 '16 at 16:06
Riemann-bitcoin.
336111
336111
This is because if $yin Cl(A)$ iff every open set containing $y$ intersects $A$.
– Riemann-bitcoin.
Oct 26 '16 at 16:10
I think I have a confusion with the definition. How I know it is that $x$ is a limit point of a subset $S$ of a topological space if every neigbourhood of $x$ intersects with $S$. So in our case if $x$ has a neigbourhood only intersecting $AB$ and another only intersecting $BA$ it should be a limit point only of $AcupB$ and not of A or B.Am I using wrond definition or saying something wrong?
– TheGeometer
Oct 26 '16 at 16:19
The $Cl(A)$ is the limit points unioned with the set A. So $Cl(A)=A cup A'$. I believe this is the confusion, though I am slightly confused on what your counterexample is saying.
– Riemann-bitcoin.
Oct 26 '16 at 16:57
add a comment |
This is because if $yin Cl(A)$ iff every open set containing $y$ intersects $A$.
– Riemann-bitcoin.
Oct 26 '16 at 16:10
I think I have a confusion with the definition. How I know it is that $x$ is a limit point of a subset $S$ of a topological space if every neigbourhood of $x$ intersects with $S$. So in our case if $x$ has a neigbourhood only intersecting $AB$ and another only intersecting $BA$ it should be a limit point only of $AcupB$ and not of A or B.Am I using wrond definition or saying something wrong?
– TheGeometer
Oct 26 '16 at 16:19
The $Cl(A)$ is the limit points unioned with the set A. So $Cl(A)=A cup A'$. I believe this is the confusion, though I am slightly confused on what your counterexample is saying.
– Riemann-bitcoin.
Oct 26 '16 at 16:57
This is because if $yin Cl(A)$ iff every open set containing $y$ intersects $A$.
– Riemann-bitcoin.
Oct 26 '16 at 16:10
This is because if $yin Cl(A)$ iff every open set containing $y$ intersects $A$.
– Riemann-bitcoin.
Oct 26 '16 at 16:10
I think I have a confusion with the definition. How I know it is that $x$ is a limit point of a subset $S$ of a topological space if every neigbourhood of $x$ intersects with $S$. So in our case if $x$ has a neigbourhood only intersecting $AB$ and another only intersecting $BA$ it should be a limit point only of $AcupB$ and not of A or B.Am I using wrond definition or saying something wrong?
– TheGeometer
Oct 26 '16 at 16:19
I think I have a confusion with the definition. How I know it is that $x$ is a limit point of a subset $S$ of a topological space if every neigbourhood of $x$ intersects with $S$. So in our case if $x$ has a neigbourhood only intersecting $AB$ and another only intersecting $BA$ it should be a limit point only of $AcupB$ and not of A or B.Am I using wrond definition or saying something wrong?
– TheGeometer
Oct 26 '16 at 16:19
The $Cl(A)$ is the limit points unioned with the set A. So $Cl(A)=A cup A'$. I believe this is the confusion, though I am slightly confused on what your counterexample is saying.
– Riemann-bitcoin.
Oct 26 '16 at 16:57
The $Cl(A)$ is the limit points unioned with the set A. So $Cl(A)=A cup A'$. I believe this is the confusion, though I am slightly confused on what your counterexample is saying.
– Riemann-bitcoin.
Oct 26 '16 at 16:57
add a comment |
Infinite point sequence from (A ∪ B) contains an infinite subsequence from A or contains an infinite subsequence from B
answered Oct 26 '16 at 16:00
kotomord
1,460626
ok but what is wrong wit my counterexample?
– TheGeometer
Oct 26 '16 at 16:04
add a comment |
Infinite point sequence from (A ∪ B) contains an infinite subsequence from A or contains an infinite subsequence from B
answered Oct 26 '16 at 16:00
kotomord
1,460626
ok but what is wrong wit my counterexample?
– TheGeometer
Oct 26 '16 at 16:04
add a comment |
Infinite point sequence from (A ∪ B) contains an infinite subsequence from A or contains an infinite subsequence from B
answered Oct 26 '16 at 16:00
kotomord
1,460626
Infinite point sequence from (A ∪ B) contains an infinite subsequence from A or contains an infinite subsequence from B
answered Oct 26 '16 at 16:00
kotomord
1,460626
answered Oct 26 '16 at 16:00
kotomord
1,460626
answered Oct 26 '16 at 16:00
kotomord
1,460626
answered Oct 26 '16 at 16:00
kotomord
1,460626
1,460626
ok but what is wrong wit my counterexample?
– TheGeometer
Oct 26 '16 at 16:04
add a comment |
ok but what is wrong wit my counterexample?
– TheGeometer
Oct 26 '16 at 16:04
ok but what is wrong wit my counterexample?
– TheGeometer
Oct 26 '16 at 16:04
ok but what is wrong wit my counterexample?
– TheGeometer
Oct 26 '16 at 16:04
add a comment |
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Choose some limit point of LHS and observe that it belong to the RHS.
– Masacroso
Oct 26 '16 at 16:06