testing continuity via filters

Proposition 1.

Let $X\mathrm{,}Y$ be topological spaces. Then a function $f\mathrm{:}X\mathrm{\to }Y$ is continuous iff it sends converging filters to converging filters.

Proof.

Suppose first $f$ is continuous. Let $𝔽$ be a filter in $X$ converging to $x$. We want to show that $f(𝔽):=\{f(F)\mid F\in 𝔽\}$ converges to $f(x)$. Let $N$ be a neighborhood of $f(x)$. So there is an open set $U$ such that $f(x)\in U\subseteq N$. So $f^{-1}(U)$ is open and contains $x$, which means that $f^{-1}(U)\in 𝔽$ by assumption. This means that $f{f}^{-1}(U)\in f(𝔽)$. Since $f{f}^{-1}(U)\subseteq U\subseteq N$, we see that $N\in f(𝔽)$ as well.

Conversely, suppose $f$ preserves converging filters. Let $V$ be an open set in $Y$ containing $f(x)$. We want to find an open set $U$ in $X$ containing $x$, such that $f(U)\subseteq V$. Let $𝔽$ be the neighborhood filter of $x$. So $𝔽\to x$. By assumption, $f(𝔽)\to f(x)$. Since $V$ is an open neighborhood of $f(x)$, we have $V\in f(𝔽)$, or $f(F)\subseteq V$ for some $F\in 𝔽$. Since $F$ is a neighborhood of $x$, it contains an open neighborhood $U$ of $x$. Furthermore, $f(U)\subseteq f(F)\subseteq V$. Since $x$ is arbitrary, $f$ is continuous.∎