testing for continuity via nets

Proposition 1.

Let $X\mathrm{,}Y$ be topological spaces and $f\mathrm{:}X\mathrm{\to }Y$. Then the following are equivalent:

1. $f$ is continuous;

2. If $(x_i)$ is a net in $X$ converging to $x$, then $(f(x_i))$ is a net in $Y$ converging to $f(x)$.

3. Whenever two nets $(x_i)$ and $(y_j)$ in $X$ converge to the same point, then $(f(x_i))$ and $(f(y_j))$converge to the same point in $Y$.

Proof.

$(1)\iff (2)$. Let $A$ be the (directed) index set for $i$. Suppose $f(x)\in U$ is open in $Y$. Then $x\in f^{-1}(U)$ is open in $X$ since $f$ is continuous. By assumption, $(x_i)$ is a net, so there is $b\in A$ such that $x_j\in f^{-1}(U)$ for all $j\ge b$. This means that $f(x_j)\in U$ for all $i\ge b$, so $(f(x_i))$ is a net too.

Conversely, suppose $f$ is not continuous, say, at a point $x\in X$. Then there is an open set $V$ containing $f(x)$ such that $f^{-1}(V)$ does not contain any open set containing $x$. Let $A$ be the set of all open sets containing $x$. Then under reverse inclusion, $A$ is a directed set (if $U_1,U_2\in A$, then $U_1\cap U_2\in A$). Define a relation $R\subseteq A\times X$ as follows:

$$(U,x)\in R\mathit{\hspace{1em}\hspace{1em}}\text{iff}\mathit{\hspace{1em}\hspace{1em}}x\in U-f^{-1}(V).$$

Then for each $U\in A$, there is an $x\in X$ such that $(U,x)\in R$, since $U⊈f^{-1}(V)$. By the axiom of choice, we get a function $d\subseteq R$ from $A$ to $X$. Write $d(U):=x_U$. Since $A$ is directed, $(x_U)$ is a net. In addition, $(x_U)$ converges to $x$ (just pick any $U\in A$, then for any $W\ge U$, we have $x\in W$ by the definition of $A$). However, $(f(x_U))$ does not converge to $f(x)$, since $x_U\notin f^{-1}(V)$ for any $U\in A$.

$(2)\iff (3)$. Suppose nets $(x_i)$ and $(y_j)$ both converge to $z\in X$. Then, by assumption, $(f(x_i))$ and $(f(y_j))$ are nets converging to $f(z)\in Y$.

Conversely, suppose $(x_i)$ converges to $x$, and $i$ is indexed by a directed set $A$. Define a net $(y_i)$ such that $y_i=x$ for all $i\in A$. Then $(y_i)=(x)$ clearly converges to $x$. Hence both $(f(x_i))$ and $(f(y_i))$ converge to the same point in $Y$. But $(f(y_i))=(f(x))$ converges to $f(x)$, we see that $(f(x_i))$ converges to $f(x)$ as well.∎

Remark. In particular, if $X,Y$ are first countable, we may replace nets by sequences in the proposition. In other words, $f$ is continuous iff it preserves converging sequences.