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Nontechnical(!) Proof: $\overline{A}^X\subseteq S\Rightarrow\overline{A}^S=\overline{A}^X$ ...while $A\subseteq S\subseteq X$
Background: $\overline{A}^S\subseteq S\nRightarrow\overline{A}^S=\overline{A}^X$ ...while $A\subseteq S\subseteq X$

Moreover, just for fun, can u proof: $\overline{A}^S\subseteq\overline{A}^X$ ...while $A\subseteq S\subseteq X$
(I did that, but using construction of closure as smallest closed set containing it.)

I need this for parts of the construction of partition of unit...

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  • $\begingroup$ What is the question? Do you want a proof of the fact that if the closure of $A$ is contained in $S$, then the relative closure is equal to the closure? $\endgroup$ – Daniel Fischer Mar 28 '14 at 15:50
  • $\begingroup$ Yes, but one that is nice and short $\endgroup$ – C-Star-W-Star Mar 28 '14 at 15:52
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Proposition: Let $X$ a topological space, and $A \subset S \subset X$. Then $\overline{A}^S = S\cap \overline{A}^X$.

Proof: $S\cap \overline{A}^X$ is a relatively closed subset of $S$ containing $A$, hence $\overline{A}^S \subset S\cap \overline{A}^X$. On the other hand, $\overline{A}^S$ is a relatively closed subset of $S$, hence there is a closed (in $X$) subset $F$ with $\overline{A}^S = S\cap F$. Since $A\subset \overline{A}^S$, it follows that $A\subset F$ and hence $\overline{A}^X\subset F$, whence $S\cap \overline{A}^X \subset S\cap F = \overline{A}^S$.

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  • $\begingroup$ Nice, thats precisely what I meant by nontechnical - thx ;) $\endgroup$ – C-Star-W-Star Mar 28 '14 at 20:24
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Let $c\in\overline{A}^X$, then any open set $U\ni c$ in $X$ intersects $A$ because of properties of closure.

$U\cap S$ is open in $S$ and intersects $A$, because $S$ is a subspace of $X$. But $c\in U\cap S$, because $\overline{A}^X\subseteq S$. This means $c\in\overline{A}^S$.

So we have $\overline{A}^X\subseteq\overline{A}^S$. The other inclusion is trivial, because $\overline{A}^X$ is a set closed in $S$ containing $A$.

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Here is a proof using relative neighborhoods:

Claim: If $A\subseteq B$ then $\overline A^B=\overline A\cap B$
Proof: Assume that $x\in\overline A^B$. If $U$ is a neighborhood of $x$, then $U\cap B$ is a relative neighborhood, so $U\cap B\cap A\ne\emptyset$, thus $x\in\overline A$.
Conversely, assume that $x\in\overline A\cap B$. If $U$ is a relative neighborhood of $x$ in $B$, then $U=B\cap U'$ for a neighborhood $U'$. Then $U'\cap A=U'\cap A\cap B=U\cap A\ne\emptyset$, so $x\in\overline A^B$.

Remark: If $B$ is open, then the claim holds even if $A$ is not a subset of $B$.

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