Prove that If B is open, then $overline{A} cap B subset overline{A cap B}$
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Let $(X,d)$ be a metric space and let $A, B subset X$. Prove that:
If B is open, then $$overline{A} cap B subset overline{A cap B}$$ where $overline{S}$ indicates closure for some set $S$.
For proof I was only able to change the prove part slightly(not sure if this is a correct start):
Since $B subset overline{B} implies overline{A} cap overline{B} subset overline{A cap B}$
real-analysis general-topology metric-spaces
asked Dec 5 at 18:47
Pumpkin
4861417
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Let $(X,d)$ be a metric space and let $A, B subset X$. Prove that:
If B is open, then $$overline{A} cap B subset overline{A cap B}$$ where $overline{S}$ indicates closure for some set $S$.
For proof I was only able to change the prove part slightly(not sure if this is a correct start):
Since $B subset overline{B} implies overline{A} cap overline{B} subset overline{A cap B}$
real-analysis general-topology metric-spaces
asked Dec 5 at 18:47
Pumpkin
4861417
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(X,d)$ be a metric space and let $A, B subset X$. Prove that:
If B is open, then $$overline{A} cap B subset overline{A cap B}$$ where $overline{S}$ indicates closure for some set $S$.
For proof I was only able to change the prove part slightly(not sure if this is a correct start):
Since $B subset overline{B} implies overline{A} cap overline{B} subset overline{A cap B}$
real-analysis general-topology metric-spaces
asked Dec 5 at 18:47
Pumpkin
4861417
Let $(X,d)$ be a metric space and let $A, B subset X$. Prove that:
If B is open, then $$overline{A} cap B subset overline{A cap B}$$ where $overline{S}$ indicates closure for some set $S$.
For proof I was only able to change the prove part slightly(not sure if this is a correct start):
Since $B subset overline{B} implies overline{A} cap overline{B} subset overline{A cap B}$
real-analysis general-topology metric-spaces
real-analysis general-topology metric-spaces
asked Dec 5 at 18:47
Pumpkin
4861417
asked Dec 5 at 18:47
Pumpkin
4861417
asked Dec 5 at 18:47
Pumpkin
4861417
asked Dec 5 at 18:47
Pumpkin
4861417
asked Dec 5 at 18:47
Pumpkin
4861417
4861417
add a comment |
add a comment |
2 Answers
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Given a set $E subset X$ you have $x in overline E$ if and only if every neighborhood of $x$ intersects $E$.
Let $x in overline A cap B$. Then $x in B$ and every neighborhood of $x$ intersects $A$.
Let $U$ be a neighborhood of $x$. Since $x in B$ and $B$ is open, $U cap B$ is a neighborhood of $x$ too.
But every neighborhood of $x$ intersects $A$, so that $U cap B$ contains a point of $A$. If you denote this point by $z$, you find $z in U cap (A cap B)$. Thus $U$ contains a point of $A cap B$.
Since $U$ was an arbitrary neighborhood of $x$, it follows that $x in overline{A cap B}$.
Consequently $overline A cap B subset overline{A cap B}$.
answered Dec 5 at 19:01
Umberto P.
38.3k13063
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Let $xin {overline A}cap B$. Since $B$ is open, we can chose $r > 0$ $B_{r}(x)subseteq B$. Let $0 < s < r$. Then $B_s(0) cap B not=emptyset$. Since $xin{overline A}$, $B_s(x)cap A not = emptyset$. Thus $B_s(x)cap A cap B not= emptyset$.
We have just shown $xin overline{A cap B}.$
answered Dec 5 at 19:04
ncmathsadist
42.2k259102
Why you assume 0 is included in the set?
– Pumpkin
Dec 6 at 11:16
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Given a set $E subset X$ you have $x in overline E$ if and only if every neighborhood of $x$ intersects $E$.
Let $x in overline A cap B$. Then $x in B$ and every neighborhood of $x$ intersects $A$.
Let $U$ be a neighborhood of $x$. Since $x in B$ and $B$ is open, $U cap B$ is a neighborhood of $x$ too.
But every neighborhood of $x$ intersects $A$, so that $U cap B$ contains a point of $A$. If you denote this point by $z$, you find $z in U cap (A cap B)$. Thus $U$ contains a point of $A cap B$.
Since $U$ was an arbitrary neighborhood of $x$, it follows that $x in overline{A cap B}$.
Consequently $overline A cap B subset overline{A cap B}$.
answered Dec 5 at 19:01
Umberto P.
38.3k13063
add a comment |
up vote
1
down vote
accepted
Given a set $E subset X$ you have $x in overline E$ if and only if every neighborhood of $x$ intersects $E$.
Let $x in overline A cap B$. Then $x in B$ and every neighborhood of $x$ intersects $A$.
Let $U$ be a neighborhood of $x$. Since $x in B$ and $B$ is open, $U cap B$ is a neighborhood of $x$ too.
But every neighborhood of $x$ intersects $A$, so that $U cap B$ contains a point of $A$. If you denote this point by $z$, you find $z in U cap (A cap B)$. Thus $U$ contains a point of $A cap B$.
Since $U$ was an arbitrary neighborhood of $x$, it follows that $x in overline{A cap B}$.
Consequently $overline A cap B subset overline{A cap B}$.
answered Dec 5 at 19:01
Umberto P.
38.3k13063
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Given a set $E subset X$ you have $x in overline E$ if and only if every neighborhood of $x$ intersects $E$.
Let $x in overline A cap B$. Then $x in B$ and every neighborhood of $x$ intersects $A$.
Let $U$ be a neighborhood of $x$. Since $x in B$ and $B$ is open, $U cap B$ is a neighborhood of $x$ too.
But every neighborhood of $x$ intersects $A$, so that $U cap B$ contains a point of $A$. If you denote this point by $z$, you find $z in U cap (A cap B)$. Thus $U$ contains a point of $A cap B$.
Since $U$ was an arbitrary neighborhood of $x$, it follows that $x in overline{A cap B}$.
Consequently $overline A cap B subset overline{A cap B}$.
answered Dec 5 at 19:01
Umberto P.
38.3k13063
Given a set $E subset X$ you have $x in overline E$ if and only if every neighborhood of $x$ intersects $E$.
Let $x in overline A cap B$. Then $x in B$ and every neighborhood of $x$ intersects $A$.
Let $U$ be a neighborhood of $x$. Since $x in B$ and $B$ is open, $U cap B$ is a neighborhood of $x$ too.
But every neighborhood of $x$ intersects $A$, so that $U cap B$ contains a point of $A$. If you denote this point by $z$, you find $z in U cap (A cap B)$. Thus $U$ contains a point of $A cap B$.
Since $U$ was an arbitrary neighborhood of $x$, it follows that $x in overline{A cap B}$.
Consequently $overline A cap B subset overline{A cap B}$.
answered Dec 5 at 19:01
Umberto P.
38.3k13063
answered Dec 5 at 19:01
Umberto P.
38.3k13063
answered Dec 5 at 19:01
Umberto P.
38.3k13063
answered Dec 5 at 19:01
Umberto P.
38.3k13063
38.3k13063
add a comment |
add a comment |
up vote
0
down vote
Let $xin {overline A}cap B$. Since $B$ is open, we can chose $r > 0$ $B_{r}(x)subseteq B$. Let $0 < s < r$. Then $B_s(0) cap B not=emptyset$. Since $xin{overline A}$, $B_s(x)cap A not = emptyset$. Thus $B_s(x)cap A cap B not= emptyset$.
We have just shown $xin overline{A cap B}.$
answered Dec 5 at 19:04
ncmathsadist
42.2k259102
Why you assume 0 is included in the set?
– Pumpkin
Dec 6 at 11:16
add a comment |
up vote
0
down vote
Let $xin {overline A}cap B$. Since $B$ is open, we can chose $r > 0$ $B_{r}(x)subseteq B$. Let $0 < s < r$. Then $B_s(0) cap B not=emptyset$. Since $xin{overline A}$, $B_s(x)cap A not = emptyset$. Thus $B_s(x)cap A cap B not= emptyset$.
We have just shown $xin overline{A cap B}.$
answered Dec 5 at 19:04
ncmathsadist
42.2k259102
Why you assume 0 is included in the set?
– Pumpkin
Dec 6 at 11:16
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $xin {overline A}cap B$. Since $B$ is open, we can chose $r > 0$ $B_{r}(x)subseteq B$. Let $0 < s < r$. Then $B_s(0) cap B not=emptyset$. Since $xin{overline A}$, $B_s(x)cap A not = emptyset$. Thus $B_s(x)cap A cap B not= emptyset$.
We have just shown $xin overline{A cap B}.$
answered Dec 5 at 19:04
ncmathsadist
42.2k259102
Let $xin {overline A}cap B$. Since $B$ is open, we can chose $r > 0$ $B_{r}(x)subseteq B$. Let $0 < s < r$. Then $B_s(0) cap B not=emptyset$. Since $xin{overline A}$, $B_s(x)cap A not = emptyset$. Thus $B_s(x)cap A cap B not= emptyset$.
We have just shown $xin overline{A cap B}.$
answered Dec 5 at 19:04
ncmathsadist
42.2k259102
answered Dec 5 at 19:04
ncmathsadist
42.2k259102
answered Dec 5 at 19:04
ncmathsadist
42.2k259102
answered Dec 5 at 19:04
ncmathsadist
42.2k259102
42.2k259102
Why you assume 0 is included in the set?
– Pumpkin
Dec 6 at 11:16
add a comment |
Why you assume 0 is included in the set?
– Pumpkin
Dec 6 at 11:16
Why you assume 0 is included in the set?
– Pumpkin
Dec 6 at 11:16
Why you assume 0 is included in the set?
– Pumpkin
Dec 6 at 11:16
add a comment |
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