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Let $(X, \tau)$ be a topological space, and let $\overline{ (-) } : \mathcal{P}(X) \to \mathcal{P}(X)$ be defined as

$$ \overline{S} := \{x : \exists(x_\alpha)_{\alpha \in \Lambda} \subseteq S \text{ net s.t. } x_\alpha \to x\}. $$

I'm trying to prove that $\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$, since I've already seen the other inclusion.

Let $x \in \overline{A \cup B}$ so that we have $(x_\alpha)_{\alpha \in \Lambda} \subseteq A \cup B$ with $x_\alpha \to x$. My idea was to adapt a similar proof for sequences, where one can assume there are infinite terms of a sequence in one of the sets, and take an appropriate subsequence. I have first proved that $x \in A$ or $x \in B$ for $\Lambda$ finite, and now am tackling the infinite case.

Concretely, now we can assume that infinitely elements of (the image of) the net are on $A$. I want to prove that $\Gamma = \{\alpha \in \Lambda : x_\alpha \in A \}$ and the inclusion $\Gamma \hookrightarrow \Lambda$ will give a subnet of the original, thus concluding that $x \in \overline{A}$. However, that does not seem immediate (or correct, even) if we do not assume $\Gamma$ cofinal, which will not happen always.

Would you mind providing any hints on whether I am on the right track, and if so on how can I finish my argument? Thanks in advance.

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The best idea is to assume $x \in \overline{A \cup B}$ giving us a net $n: \Lambda \to X$ such that $n \to x$ and for all $\alpha \in \Lambda$ we have $n(\alpha) \in A \cup B$, and then define $\Lambda(A) = \{ \alpha \in \Lambda: n(\alpha) \in A\}$ and likewise $\Lambda(B)$ and by definition we have that $\Lambda = \Lambda(A) \cup \Lambda(B)$, (not disjoint necessarily, of course, but that's OK).

Can it be true that neither is cofinal in $\Lambda$? (Hint : no) and then you're done (a cofinal subset gives another (sub)net using inclusion, as you suggested ) What does it mean to be not cofinal?? Get a contradiction from the definition of a directed set.

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  • $\begingroup$ This is enlightening! I will read this carefully and rewrite my proof when I'm less tired, and accept the answer accordingly. Thanks so much for the insight :) $\endgroup$ – Guido A. Sep 8 '18 at 8:51

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