Constructing a net in a topological space.

Question

Let $X$ be a topological space , and $S \subseteq X$ be a nonempty set. Pick $x \in S$. Can we always construct a net $(x_i)_{i \in I}$ in $X$ such that $x_i \to x $ , and $$ S = \{ x_i ~| ~ i \in I \} $$

If so, is it possible to choose the index (directed) set $I$ a totally ordered set?

I think the answer is yes, because any nonempty set can be totally ordered, so we can put a total order on $S$, and take $I=S$ but I dont know how we can modify this order a little bit in order the convergence part is satisfied !

Answer 1

The net is quite easy: take any linear order $<$ on $S\setminus\{x\}$ (needs AC in general) and extend the linear order to $S$ by letting $x$ be larger than all other points, adding a maximum at the end.

Then trivial map $(S,<) \to X$, sending $s$ to $s$ is a net and converges to $x$ as $\{x\}$ is by itself cofinal.

Answer 2

More generally, if $x \in \overline{S}$, then the usual construction of a net in $S$ converging to $x$ gives a net with image equal to $S$. To make things concrete, the net has index set $I$ whose elements are $(U, y)$ where $U$ is an open neighborhood of $x$ and $y \in S \cap U$ (with preorder defined so that $(U, y) \le (V, z)$ exactly when $V \subseteq U$), and the net is $(U, y) \mapsto y$. Then for any $y \in S$, $(X, y) \in I$ and $x_{(X, y)} = y$.