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In proving that the equivalence between different definitions of continuous functions between topological spaces, I come across this lemma. Could you have a check if my attempt is correct?

Lemma: Let $X$ be a topological space and $A \subseteq X$. Then $\overline A$ is the set of limits of convergent nets with values in $A$.

Proof: For $x \in X$, let $\mathcal N_x$ be the set of all neighborhoods of $x$. We have $$\overline A = \{x \in X \mid \forall U \in \mathcal N_x: U \cap A \neq \emptyset \}.$$

Then we define a pre-order $\le$ on $\mathcal N_x$ by $U_1 \le U_2 \iff U_2 \subseteq U_1$. Then $\mathcal N_x$ is a directed set. Let $x \in \overline A$. By axiom of choice, for each $U \in \mathcal N_x$, there is $x_U \in U \cap A$. Hence $(x_U)_{U \in \mathcal N_x}$ is a net in $A$ such that $x_U \to x$. Conversely, let $(x_d)_{d\in D}$ be a net in $A$ that converges to $x \in X$. For each $U \in \mathcal N_x$, there is $d' \in D$ such that $x_d \in U$ for all $d \ge d'$. This means $U \cap A \neq \emptyset$ for all $U \in \mathcal N_x$. Hence $x \in \overline A$.

Henno Brandsma
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Akira
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1 Answers1

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Good for mentioning the rôle of AC. Explain why $\mathcal N_x$ is directed under $\le$ (yes, intersection of neighbourhoods ...). Add also the argument why $(x_U)_{U \in \mathcal N_x}$ converges to $x$ (it's a minor thing but shows why we have to define the pre-order the way we do). It depends on the intended audience, I suppose, but in home work or self-study fill in all the details...

The reverse direction I have no quarrel with, as remarking that $x_{d'} \in A \cap U$ to witness the non-emptyness, might be too trivial? I'll let it pass...

Ideas are fine.

Henno Brandsma
  • 242,131