Is it true that the intersection of the closures of sets $A$ and $ B$ is equal to the closure of their...
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Is it true that the intersection of the closures of sets $A$ and $B$ is equal to the closure of their intersection?
$ cl(A)cap{cl(B)}=cl(Acap{B})$ ?
general-topology
edited Dec 19 '18 at 7:24
dmtri
1,4692521
asked Dec 19 '18 at 6:39
Zhaniya ShaimakhanovaZhaniya Shaimakhanova
111
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closed as off-topic by Saad, Lee David Chung Lin, José Carlos Santos, user10354138, Jyrki Lahtonen Dec 19 '18 at 13:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lee David Chung Lin, José Carlos Santos, user10354138, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Is it true that the intersection of the closures of sets $A$ and $B$ is equal to the closure of their intersection?
$ cl(A)cap{cl(B)}=cl(Acap{B})$ ?
general-topology
edited Dec 19 '18 at 7:24
dmtri
1,4692521
asked Dec 19 '18 at 6:39
Zhaniya ShaimakhanovaZhaniya Shaimakhanova
111
$endgroup$
closed as off-topic by Saad, Lee David Chung Lin, José Carlos Santos, user10354138, Jyrki Lahtonen Dec 19 '18 at 13:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lee David Chung Lin, José Carlos Santos, user10354138, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Is it true that the intersection of the closures of sets $A$ and $B$ is equal to the closure of their intersection?
$ cl(A)cap{cl(B)}=cl(Acap{B})$ ?
general-topology
edited Dec 19 '18 at 7:24
dmtri
1,4692521
asked Dec 19 '18 at 6:39
Zhaniya ShaimakhanovaZhaniya Shaimakhanova
111
$endgroup$
Is it true that the intersection of the closures of sets $A$ and $B$ is equal to the closure of their intersection?
$ cl(A)cap{cl(B)}=cl(Acap{B})$ ?
general-topology
general-topology
edited Dec 19 '18 at 7:24
dmtri
1,4692521
asked Dec 19 '18 at 6:39
Zhaniya ShaimakhanovaZhaniya Shaimakhanova
111
edited Dec 19 '18 at 7:24
dmtri
1,4692521
asked Dec 19 '18 at 6:39
Zhaniya ShaimakhanovaZhaniya Shaimakhanova
111
edited Dec 19 '18 at 7:24
dmtri
1,4692521
edited Dec 19 '18 at 7:24
dmtri
1,4692521
edited Dec 19 '18 at 7:24
dmtri
1,4692521
1,4692521
asked Dec 19 '18 at 6:39
Zhaniya ShaimakhanovaZhaniya Shaimakhanova
111
asked Dec 19 '18 at 6:39
Zhaniya ShaimakhanovaZhaniya Shaimakhanova
111
asked Dec 19 '18 at 6:39
Zhaniya ShaimakhanovaZhaniya Shaimakhanova
111
111
closed as off-topic by Saad, Lee David Chung Lin, José Carlos Santos, user10354138, Jyrki Lahtonen Dec 19 '18 at 13:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lee David Chung Lin, José Carlos Santos, user10354138, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Lee David Chung Lin, José Carlos Santos, user10354138, Jyrki Lahtonen Dec 19 '18 at 13:27
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Lee David Chung Lin, José Carlos Santos, user10354138, Jyrki Lahtonen
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
3 Answers
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No, the rationals and the irrationals (in the reals) are disjoint so $operatorname{cl}(A cap B) = operatorname{cl}{emptyset}=emptyset$ while $operatorname{cl}(A) = operatorname{cl}(B) = mathbb{R}$
answered Dec 19 '18 at 6:44
Henno BrandsmaHenno Brandsma
107k347114
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No. Take $A=(-1,0), B=(0,1)$. Note that $0$ is in the closure of both of these sets.
answered Dec 19 '18 at 6:41
Kavi Rama MurthyKavi Rama Murthy
56k42158
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No, it's not true. Look at this page, or this question.
A simple counterexample stolen from the question cited above is as follows:
Take $A = (0,1)$ and $B = (1,2)$. Then we have
$$operatorname{cl}(A) cap operatorname{cl}(B) = [0,1] cap [1,2] = lbrace 1 rbrace $$
but
$$operatorname{cl}(A cap B) = operatorname{cl}(emptyset) = emptyset$$
answered Dec 19 '18 at 6:44
bubbabubba
30.3k33086
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add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, the rationals and the irrationals (in the reals) are disjoint so $operatorname{cl}(A cap B) = operatorname{cl}{emptyset}=emptyset$ while $operatorname{cl}(A) = operatorname{cl}(B) = mathbb{R}$
answered Dec 19 '18 at 6:44
Henno BrandsmaHenno Brandsma
107k347114
$endgroup$
add a comment |
$begingroup$
No, the rationals and the irrationals (in the reals) are disjoint so $operatorname{cl}(A cap B) = operatorname{cl}{emptyset}=emptyset$ while $operatorname{cl}(A) = operatorname{cl}(B) = mathbb{R}$
answered Dec 19 '18 at 6:44
Henno BrandsmaHenno Brandsma
107k347114
$endgroup$
add a comment |
$begingroup$
No, the rationals and the irrationals (in the reals) are disjoint so $operatorname{cl}(A cap B) = operatorname{cl}{emptyset}=emptyset$ while $operatorname{cl}(A) = operatorname{cl}(B) = mathbb{R}$
answered Dec 19 '18 at 6:44
Henno BrandsmaHenno Brandsma
107k347114
$endgroup$
No, the rationals and the irrationals (in the reals) are disjoint so $operatorname{cl}(A cap B) = operatorname{cl}{emptyset}=emptyset$ while $operatorname{cl}(A) = operatorname{cl}(B) = mathbb{R}$
answered Dec 19 '18 at 6:44
Henno BrandsmaHenno Brandsma
107k347114
answered Dec 19 '18 at 6:44
Henno BrandsmaHenno Brandsma
107k347114
answered Dec 19 '18 at 6:44
Henno BrandsmaHenno Brandsma
107k347114
answered Dec 19 '18 at 6:44
Henno BrandsmaHenno Brandsma
107k347114
107k347114
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add a comment |
$begingroup$
No. Take $A=(-1,0), B=(0,1)$. Note that $0$ is in the closure of both of these sets.
answered Dec 19 '18 at 6:41
Kavi Rama MurthyKavi Rama Murthy
56k42158
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add a comment |
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No. Take $A=(-1,0), B=(0,1)$. Note that $0$ is in the closure of both of these sets.
answered Dec 19 '18 at 6:41
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
add a comment |
$begingroup$
No. Take $A=(-1,0), B=(0,1)$. Note that $0$ is in the closure of both of these sets.
answered Dec 19 '18 at 6:41
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
No. Take $A=(-1,0), B=(0,1)$. Note that $0$ is in the closure of both of these sets.
answered Dec 19 '18 at 6:41
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Dec 19 '18 at 6:41
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Dec 19 '18 at 6:41
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Dec 19 '18 at 6:41
Kavi Rama MurthyKavi Rama Murthy
56k42158
56k42158
add a comment |
add a comment |
$begingroup$
No, it's not true. Look at this page, or this question.
A simple counterexample stolen from the question cited above is as follows:
Take $A = (0,1)$ and $B = (1,2)$. Then we have
$$operatorname{cl}(A) cap operatorname{cl}(B) = [0,1] cap [1,2] = lbrace 1 rbrace $$
but
$$operatorname{cl}(A cap B) = operatorname{cl}(emptyset) = emptyset$$
answered Dec 19 '18 at 6:44
bubbabubba
30.3k33086
$endgroup$
add a comment |
$begingroup$
No, it's not true. Look at this page, or this question.
A simple counterexample stolen from the question cited above is as follows:
Take $A = (0,1)$ and $B = (1,2)$. Then we have
$$operatorname{cl}(A) cap operatorname{cl}(B) = [0,1] cap [1,2] = lbrace 1 rbrace $$
but
$$operatorname{cl}(A cap B) = operatorname{cl}(emptyset) = emptyset$$
answered Dec 19 '18 at 6:44
bubbabubba
30.3k33086
$endgroup$
add a comment |
$begingroup$
No, it's not true. Look at this page, or this question.
A simple counterexample stolen from the question cited above is as follows:
Take $A = (0,1)$ and $B = (1,2)$. Then we have
$$operatorname{cl}(A) cap operatorname{cl}(B) = [0,1] cap [1,2] = lbrace 1 rbrace $$
but
$$operatorname{cl}(A cap B) = operatorname{cl}(emptyset) = emptyset$$
answered Dec 19 '18 at 6:44
bubbabubba
30.3k33086
$endgroup$
No, it's not true. Look at this page, or this question.
A simple counterexample stolen from the question cited above is as follows:
Take $A = (0,1)$ and $B = (1,2)$. Then we have
$$operatorname{cl}(A) cap operatorname{cl}(B) = [0,1] cap [1,2] = lbrace 1 rbrace $$
but
$$operatorname{cl}(A cap B) = operatorname{cl}(emptyset) = emptyset$$
answered Dec 19 '18 at 6:44
bubbabubba
30.3k33086
edited Dec 20 '18 at 2:05
answered Dec 19 '18 at 6:44
bubbabubba
30.3k33086
answered Dec 19 '18 at 6:44
bubbabubba
30.3k33086
answered Dec 19 '18 at 6:44
bubbabubba
30.3k33086
30.3k33086
add a comment |
add a comment |
