Let (X, d) be a metric space and let A be a subset of X and O be an open subset of X. Prove that $O cap...
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The intersection of a closure with an open set
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I don't know how to use the fact that O is open. Its interior must be itself, so I guess I should take the complement of the sets in question. But I'm not sure how to proceed.
real-analysis
asked Nov 26 at 9:58
davidh
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marked as duplicate by Henno Brandsma, user126154, José Carlos Santos real-analysis
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Nov 26 at 10:06
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The intersection of a closure with an open set
3 answers
I don't know how to use the fact that O is open. Its interior must be itself, so I guess I should take the complement of the sets in question. But I'm not sure how to proceed.
real-analysis
asked Nov 26 at 9:58
davidh
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This question already has an answer here:
The intersection of a closure with an open set
3 answers
I don't know how to use the fact that O is open. Its interior must be itself, so I guess I should take the complement of the sets in question. But I'm not sure how to proceed.
real-analysis
asked Nov 26 at 9:58
davidh
695
This question already has an answer here:
The intersection of a closure with an open set
3 answers
I don't know how to use the fact that O is open. Its interior must be itself, so I guess I should take the complement of the sets in question. But I'm not sure how to proceed.
This question already has an answer here:
The intersection of a closure with an open set
3 answers
real-analysis
real-analysis
asked Nov 26 at 9:58
davidh
695
asked Nov 26 at 9:58
davidh
695
asked Nov 26 at 9:58
davidh
695
asked Nov 26 at 9:58
davidh
695
asked Nov 26 at 9:58
davidh
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1 Answer
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Suppose $x in O cap overline{A}$. This means that $x in O$ and $x in overline{A}$. Suppose $B$ is any open ball containing $x$, then $O cap B$ is also open (this is where we use $O$ is open) and contains $x$, and as $x in overline{A}$, we know that $(B cap O) cap A neq emptyset$. But this means that $B cap (O cap A) neq emptyset$ as well and as $B$ was arbitrary, $x in overline{O cap A}$, as required to prove the inclusion.
answered Nov 26 at 10:03
Henno Brandsma
104k346113
I don't really follow the statement that $(B cap O) cap A neq emptyset$. Could you explain that a bit?
– davidh
Nov 26 at 10:07
couldn't x be in closure of A but not in A?
– davidh
Nov 26 at 10:07
$x in overline{A}$ iff every open set (or open ball) that contains $x$ intersects $A$. $B cap O$ is such an open set, and we know $x in overline{A}$ from the start @davidh
– Henno Brandsma
Nov 26 at 10:09
I see. So, for the last bit, because B is arbitrary, we invoke the iff condition for closedness on $O cap A$ right?
– davidh
Nov 26 at 10:13
@davidh yes we use both directions.
– Henno Brandsma
Nov 26 at 10:14
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
Suppose $x in O cap overline{A}$. This means that $x in O$ and $x in overline{A}$. Suppose $B$ is any open ball containing $x$, then $O cap B$ is also open (this is where we use $O$ is open) and contains $x$, and as $x in overline{A}$, we know that $(B cap O) cap A neq emptyset$. But this means that $B cap (O cap A) neq emptyset$ as well and as $B$ was arbitrary, $x in overline{O cap A}$, as required to prove the inclusion.
answered Nov 26 at 10:03
Henno Brandsma
104k346113
I don't really follow the statement that $(B cap O) cap A neq emptyset$. Could you explain that a bit?
– davidh
Nov 26 at 10:07
couldn't x be in closure of A but not in A?
– davidh
Nov 26 at 10:07
$x in overline{A}$ iff every open set (or open ball) that contains $x$ intersects $A$. $B cap O$ is such an open set, and we know $x in overline{A}$ from the start @davidh
– Henno Brandsma
Nov 26 at 10:09
I see. So, for the last bit, because B is arbitrary, we invoke the iff condition for closedness on $O cap A$ right?
– davidh
Nov 26 at 10:13
@davidh yes we use both directions.
– Henno Brandsma
Nov 26 at 10:14
add a comment |
Suppose $x in O cap overline{A}$. This means that $x in O$ and $x in overline{A}$. Suppose $B$ is any open ball containing $x$, then $O cap B$ is also open (this is where we use $O$ is open) and contains $x$, and as $x in overline{A}$, we know that $(B cap O) cap A neq emptyset$. But this means that $B cap (O cap A) neq emptyset$ as well and as $B$ was arbitrary, $x in overline{O cap A}$, as required to prove the inclusion.
answered Nov 26 at 10:03
Henno Brandsma
104k346113
I don't really follow the statement that $(B cap O) cap A neq emptyset$. Could you explain that a bit?
– davidh
Nov 26 at 10:07
couldn't x be in closure of A but not in A?
– davidh
Nov 26 at 10:07
$x in overline{A}$ iff every open set (or open ball) that contains $x$ intersects $A$. $B cap O$ is such an open set, and we know $x in overline{A}$ from the start @davidh
– Henno Brandsma
Nov 26 at 10:09
I see. So, for the last bit, because B is arbitrary, we invoke the iff condition for closedness on $O cap A$ right?
– davidh
Nov 26 at 10:13
@davidh yes we use both directions.
– Henno Brandsma
Nov 26 at 10:14
add a comment |
Suppose $x in O cap overline{A}$. This means that $x in O$ and $x in overline{A}$. Suppose $B$ is any open ball containing $x$, then $O cap B$ is also open (this is where we use $O$ is open) and contains $x$, and as $x in overline{A}$, we know that $(B cap O) cap A neq emptyset$. But this means that $B cap (O cap A) neq emptyset$ as well and as $B$ was arbitrary, $x in overline{O cap A}$, as required to prove the inclusion.
answered Nov 26 at 10:03
Henno Brandsma
104k346113
Suppose $x in O cap overline{A}$. This means that $x in O$ and $x in overline{A}$. Suppose $B$ is any open ball containing $x$, then $O cap B$ is also open (this is where we use $O$ is open) and contains $x$, and as $x in overline{A}$, we know that $(B cap O) cap A neq emptyset$. But this means that $B cap (O cap A) neq emptyset$ as well and as $B$ was arbitrary, $x in overline{O cap A}$, as required to prove the inclusion.
answered Nov 26 at 10:03
Henno Brandsma
104k346113
edited Nov 26 at 10:18
answered Nov 26 at 10:03
Henno Brandsma
104k346113
answered Nov 26 at 10:03
Henno Brandsma
104k346113
answered Nov 26 at 10:03
Henno Brandsma
104k346113
104k346113
I don't really follow the statement that $(B cap O) cap A neq emptyset$. Could you explain that a bit?
– davidh
Nov 26 at 10:07
couldn't x be in closure of A but not in A?
– davidh
Nov 26 at 10:07
$x in overline{A}$ iff every open set (or open ball) that contains $x$ intersects $A$. $B cap O$ is such an open set, and we know $x in overline{A}$ from the start @davidh
– Henno Brandsma
Nov 26 at 10:09
I see. So, for the last bit, because B is arbitrary, we invoke the iff condition for closedness on $O cap A$ right?
– davidh
Nov 26 at 10:13
@davidh yes we use both directions.
– Henno Brandsma
Nov 26 at 10:14
add a comment |
I don't really follow the statement that $(B cap O) cap A neq emptyset$. Could you explain that a bit?
– davidh
Nov 26 at 10:07
couldn't x be in closure of A but not in A?
– davidh
Nov 26 at 10:07
$x in overline{A}$ iff every open set (or open ball) that contains $x$ intersects $A$. $B cap O$ is such an open set, and we know $x in overline{A}$ from the start @davidh
– Henno Brandsma
Nov 26 at 10:09
I see. So, for the last bit, because B is arbitrary, we invoke the iff condition for closedness on $O cap A$ right?
– davidh
Nov 26 at 10:13
@davidh yes we use both directions.
– Henno Brandsma
Nov 26 at 10:14
I don't really follow the statement that $(B cap O) cap A neq emptyset$. Could you explain that a bit?
– davidh
Nov 26 at 10:07
I don't really follow the statement that $(B cap O) cap A neq emptyset$. Could you explain that a bit?
– davidh
Nov 26 at 10:07
couldn't x be in closure of A but not in A?
– davidh
Nov 26 at 10:07
couldn't x be in closure of A but not in A?
– davidh
Nov 26 at 10:07
$x in overline{A}$ iff every open set (or open ball) that contains $x$ intersects $A$. $B cap O$ is such an open set, and we know $x in overline{A}$ from the start @davidh
– Henno Brandsma
Nov 26 at 10:09
$x in overline{A}$ iff every open set (or open ball) that contains $x$ intersects $A$. $B cap O$ is such an open set, and we know $x in overline{A}$ from the start @davidh
– Henno Brandsma
Nov 26 at 10:09
I see. So, for the last bit, because B is arbitrary, we invoke the iff condition for closedness on $O cap A$ right?
– davidh
Nov 26 at 10:13
I see. So, for the last bit, because B is arbitrary, we invoke the iff condition for closedness on $O cap A$ right?
– davidh
Nov 26 at 10:13
@davidh yes we use both directions.
– Henno Brandsma
Nov 26 at 10:14
@davidh yes we use both directions.
– Henno Brandsma
Nov 26 at 10:14
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