The topology induced by a "good" net convergence notion induces a net convergence notion as originally specified

Question

The following is from Problem 11D in Willard's General topology textbook.

Suppose we have some notion of convergence on a set $X$ satisfying the following properties. Fix $x\in X$ and let $I$ be a directed set.

(a) If $x_i=x$ for each $i\in I$, then the net $(x_i)$ converges to $x$.

(b) If $(x_i)$ converges to $x$, then every subnet of $(x_i)$ converges to $x$.

(c) If every subnet of $(x_i)$ has a subnet converging to $x$, then $(x_i)$ converges to $x$.

(d) (Diagonal principle) If $(x_i)$ converges to $x$ and, for each $i\in I$, a net $(x^i_j)_{j\in J_i}$ converges to $x_i$, then there is a diagonal net converging to $x$; i.e., the net $(x^i_j)_{i\in I,\,j\in J_i}$, ordered lexicographically by $I$, then by $J_i$, has a subnet which converges to $x$.

Then if the closure of a subset $E$ of $X$ is defined by $$ \overline{E}:=\{x\in X \mid x_i\to x\ \text{for some net $(x_i)$ contained in $E$}\}, $$ the result is a topological space in which the notion of net convergence is as originally specified.

I'm struggling to show that the result is a topological space in which the notion of net convergence is as originally specified. It's pretty simple to show that, if a net $(x_i)_{i\in I}$ converges to $x$ in the original notion, then it also converges to $x$ in the topology induced by this original notion. Now, suppose such a net converges to $x$ in the induced topology. How do I show that it yet converges in the original notion?

Since the original notion is given "by imposition", with no restrictions other than (a), (b), (c) and (d), at first sight it seems "impossible" to go back and guarantee the net converges in the original sense, at least, I haven't had any idea.

For example:

• a direct proof seems really not to exist, because there is not such an explicit condition $\mathscr{C}=\mathscr{C}((x_i)_{i\in I},x)$ for which if $\scr C$ is verified for $(x_i)_{i\in I}$ and $x$, then $x_i\to x$ in the original sense;
• say we try to prove it by contradiction. Suppose $x_i\not\to x$ in the original sense. This seems to lack information just like above! Well, I still tried to go on: if $x_i\not\to x$, then the set $A=\{x_i\,:\, i\in I\}-\{x\}$ is not closed in the result topology, since $x\notin A$ and there is an obvious net on $A$ which converges (in the result topology, by hypothesis) to $x$. Then what?

Edit: Indeed, the "obvious net on $A$" is not that obvious. Since we are taking $x$ out of $\{x_i:i\in I\}$, maybe we cannot consider the whole net and have to exclude some indexes $i\in I$ in order to get a net on A.

This question is related but not equal to this one.

Answer 1

If $(x_i) \to x$ in the induced topology, suppose it did not converge in the original convergence notion. This means by (c) that there is a subnet $(y_j) ( j \in J)$ of this net, such that no subnet of that $(y_j)_j$ converges to $x$ for this convergence notion. For every $j$, the $x$ lies in the closure of $T_j:=\{y_k: k \ge j\}$ (in the induced topology, because of the convergence of $(y_j)_j$ to $x$ in the induced topology!) so there is a net $(z^j)_{j \in I_j}$ from $T_j$ that converges to $x$ in the convergence notion (by the definition of the closure!). But then property (d), (applied to a constant net $x$) gives us a subnet of $(y_j)$ that converges to $x$ (this is some tedious checking of definitions), which is a contradiction with how the net $(y_j)$ was chosen in the first place.