# Nets, dense subsets and continuous maps

Let $X$ and $Y$ be topological spaces, with $Y$ regular. Consider a dense subset $D\subset X$, a continuous map $f:D\rightarrow Y$, and a map $g:X\rightarrow Y$ (i.e. $g$ is not assumed continuous). Suppose that for any net $\{d_{\alpha}\}$ in $D$ with $d_{\alpha}\to x\in X$ we have $f(d_{\alpha})\to g(x)$.

I am trying to show that the the map $\mathbf{g}$ is continuous.

• What is $H$?  – Stefan Hamcke Aug 4 '13 at 14:43
• I think he/she meant $D$. – John Aug 4 '13 at 15:03

Since $$f$$ is continuous, your assumptions imply that $$f=g$$ on the entire $$X$$. Hence, $$g$$ is continuous.

Edit: As John correctly noted, the above sentence is insufficient since $$f$$ is only defined on $$E$$. Here is the detailed proof of continuity of $$g$$ on $$X$$, not merely on $$D$$.

First, it is clear from continuity of $$f: D\to Y$$ and the hypothesis, that $$f=g$$ on $$D$$. The statement now follows from the following:

Lemma. Suppose that $$Y$$ is regular. Let $$E\subset X$$ be a dense subset of a topological space $$X$$ and let $$f: X\to Y$$ be a map such that for every $$x\in X$$ and every net $$(e_\gamma)$$ in $$E$$ converging to $$x$$, we have $$f(x)\in \lim_\gamma f(e_\gamma).$$ ($$Y$$ need not be Hausdorff, so $$\lim_\gamma y_\gamma$$ is the set of all limits of the net $$(y_\gamma)$$ in $$Y$$.) Then $$f$$ is continuous on $$X$$.

Proof. Let $$x\in X$$, $$(x_\alpha)$$ be a net in $$X$$ converging to $$x$$. Suppose that the net $$y_\alpha=f(x_\alpha)$$ does not converge to $$y=f(x)$$. Then, without loss of generality, after passing to a subnet, we can assume that the closure $$A$$ of the (image of the) net $$(y_\alpha)$$ in $$Y$$ does not contain $$y$$. By regularity of $$Y$$, there exists a pair of disjoint open sets $$U$$ and $$V$$ with $$y\in U$$, $$A\subset V$$.

Since $$E$$ is dense in $$X$$, for each $$x_\alpha\in X$$, there exists a net $$(e_{\alpha,\beta})$$ in $$E$$ converging to $$x_{\alpha}$$. By assumption, $$f(x_\alpha)= \lim_\beta f(e_{\alpha,\beta}).$$ Now, use the "standard" diagonal net argument (which you probably have seen many times in the case of sequences), there exists a net $$(e_\gamma)$$ in $$E$$ which converges to $$x$$ and whose elements are all of the form $$e_{\alpha,\beta}$$. Hence, by the assumption, the net $$f(e_\gamma)$$ converges to $$y=f(x)$$.

On the other hand, since $$\lim_\beta f(e_{\alpha,\beta})=y_\alpha \in V,$$ we may choose the net $$(e_\gamma)$$ so that $$f(e_\gamma)\in V$$ for all $$\gamma$$ in the directed set. Contradiction with the fact that $$f(e_\gamma)$$ converges to $$y\in U$$ and $$U$$ is disjoint from $$V$$. qed

• $f=g$ on $X$? $f$'s domain is $D$. What do you mean? $g$ seems to extend $f$. So I would say that $f=g$ on $D$. But that it does so continuously is not so clear to me. – John Aug 4 '13 at 15:01
• @John: You are right, of course, I was careless. I now wrote a detailed proof. – Moishe Kohan Aug 5 '13 at 2:34