Closure and Convergent Nets

Question

Can anyone help me prove the following statement:

For a topological space $X$ and a subspace $Y$, prove that $x \in X$ belongs to the closure of $Y$ $\Rightarrow$ there is a directed set $A$ and a convergent net $f:A \rightarrow Y$ such that $x=\lim_{a \in A}f(a)$.

Many thanks.

Answer 1

Assume $A$ is a subset of a space $S$.
If $n$ is a net into $A$ that converges to $x$,
then $ x \in \overline{A}$. Proof is direct.

If $x \in \overline{A}$, then construct
a net $n$ into $A$ that converges to $x$.
For all open neighborhoods $U \ni x,$ exists $a_U$ in A.
Order $N,$ the open neighborhoods of $x,$ by reverse inclusion.
If $U,V \in N$, then $U \cap V \subseteq U,V$ directs $N$.
Show $n:N \rightarrow A$, $U \mapsto a_U$ is a net into $A$ converging to $x$.

Answer 2

Let $A$ be the directed set $\mathcal{N}_x$, the set of all neighbourhoods of $x$, with direction $U \ge V$ iff $U \subseteq V$ (directedness is clear, because $V,W \in A$ then $V \cap W \in A$ and $V \cap W \ge V, V \cap W \ge W$) and define the net $f:A\to Y$ by picking some $f(N)\in N \cap Y$ for each $N \in \mathcal{N}_x$. The existence of such $f(N)$ follows from $x \in \overline{Y}$: every neighbourhood of $x$ must intersect $Y$. By the axiom of choice we can pick a point from all these non-empty intersections.

This net clearly converges to $x$, almost by definition/construction: Let $O$ be any open set containing $x$, then by definition $O \in \mathcal{N}_x =A$ and if $A \ge O$ we know that $f(A) \in A \subseteq O$, so the tail of the net starting from $O$ is contained in $O$, as required.

This illustrates that allowing directed sets as index sets assures us of lots of relevant nets, while sequences need not always exist; countability is special.