Prove $A \subset B$ implies $\overline A \subset B$ when B is closed

Asked Nov 10 '21 at 01:02

Active Nov 10 '21 at 01:02

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Say I have subsets A and B of a metric space $(M,d)$ and $A \subset B$

I was wondering when/how we could use the above information to show $\overline A \subset B$.

I know that $A \subset B$ implies $\overline A \subset \overline B$. I have seen several examples of open subsets (Say A = (0,1) and B= (0,1] in the Real Numbers) that disprove the idea I'm wondering about on this site.

In that case, say $A \subset B$ and $B$ is closed.

If the above conditions are true, can it be proven that $\overline A \subset B$?

asked Nov 10 '21 at 01:02

Bob Mcdonald

3

$\bar A$ is the smallest closed set containing $A$ (i.e. the intersection of all closed sets that contain $A$). Since $B$ is one of those sets, $\bar A\subset B$. – anthonyquas Nov 10 '21 at 01:04
It depends on the definition of closure. – copper.hat Nov 10 '21 at 01:05
I found your question here https://math.stackexchange.com/q/121236/840578 – david hernandez Nov 10 '21 at 01:06
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Alternatively, note that if $B$ is closed, $B = \overline B$, thus $A \subset B$ implies $\overline A \subset \overline B = B$. – Rushy Nov 10 '21 at 01:08
@Rushy That answers the questions about closure that had popped up from david hernandez's comment. By definition of closure, your comment is true for all subsets of metric spaces? – Bob Mcdonald Nov 10 '21 at 01:10
It's true for any topological space. – Rushy Nov 10 '21 at 09:25

Prove $A \subset B$ implies $\overline A \subset B$ when B is closed

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