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Let $X$ and $Y$ be topological spaces (not necessarily assumed to be Hausdorff or to have any additional property) and let $f:X\to Y$ be a given function. Is it true that $f$ is continuous iff for every $x\in X$, and every net $\{x_i\}_i$ converging to $x$, one has that $f(x_i)\to f(x)$.

PS: I have searched for this specific question in MSE and, although I found several posts discussing it (this, this, and this) with varying degress of objectivity, and using various additional hypothesis (Hausdorff, completely regular, first countable) I do not believe it can be found in the exact terms above.

I therefore thought it would be nice to register it here.

I am also providing an answer which I hope is similar to the one in The Book $\ddot \smile$

Ruy
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2 Answers2

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The proof of the "only if" part is well known, so let us prove the "if" part.

Assuming by contradiction that $f$ is discontinuous at a point $x$ in $X$, we will build a net $\{x_i\}_{i\in I}$ converging to $x$, such that $\{f(x_i)\}_{i\in I}$ does not converge to $f(x)$. The set of indices for our net will be the set $N_x$ formed by all neighborhoods of $x$, and viewed as a directed set with order given by reverse inclusion, namely, $$ U\geq V \Leftrightarrow U\subseteq V. $$

Since $f$ is discontinuous at $x$, there exists a neighborhood $U$ of $f(x)$ such that $f(V)\not\subseteq U$, for all $V\in N_x$. Therefore, for any such $V$ we may choose some $x_V\in V$ such that $f(x_V)\not\in U$.

It is then evident that the net $\{x_V\}_{V\in N_x}$ converges to $x$, while $\{f(x_V)\}_{V\in N_x}$ does not converge to $f(x)$.

PS: The same argument above shows that $f$ is continuous at a given point $x_0$, if and only if, for every net $\{x_i\}_{i\in I}$, converging to $x_0$, one has that $\{f(x_i)\}_{i\in I}$ converges to $f(x_0)$.

Ruy
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  • I think everyone that knows what a net is knows this result. – J. De Ro Dec 26 '20 at 14:16
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    @QuantumSpace, I agree but I think there is a lot of people unsure of the precise hypotheses (namely none) needed to prove it. Just check out the posts I refered to above to see what I mean. – Ruy Dec 26 '20 at 14:22
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    A bit strange I would say, but yes that makes your post useful for the community! I'd say keep up the good work :) – J. De Ro Dec 26 '20 at 14:27
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Suppose first that $f$ is continuous. Let $(x_\alpha)_{\alpha \in A}$ converge to $x$. Let $V \subset Y$ be open containing $f(x)$. Then $$x \in f^{-1}(V)$$ which is open by continuity of $f$, thus there exists some $\alpha_0 \in A$ such that $$x_\alpha \in f^{-1}(V)$$ for all $\alpha \ge \alpha_0$. Then we have $$f(x_\alpha) \in V$$ forcing $f(x_\alpha) \rightarrow f(x)$.

Conversely, suppose that if $(x_\alpha)_{\alpha \in A}$ converges to $x$, then $(f(x_\alpha))_{\alpha \in A}$ converges to $f(x)$. And assume $f$ is discontinuous at $x$. I.e., there exists some open set $V \subset Y$ containing $f(x)$ such that $$x \in (f^{-1}(V))^\circ = \overline{f^{-1}(V^c)}.$$ Then one can conclude that there exists a net $(x_\alpha)$ such that for all $\alpha \geq \alpha_0$ one has $$x_\alpha \in f^{-1}(V^c).$$ Then for all $\alpha \geq \alpha_0$ one has that since $f(x) \in V$ that $$f(x_\alpha) \in V.$$ This is a contradiction since $f(x_\alpha) \in V^c$ thus $f$ is continuous.

homosapien
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