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Hometesting continuity via filters

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# testing continuity via filters

###### Proposition 1.

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

###### Proof.

Suppose first $f$ is continuous. Let $\mathbb{F}$ be a filter in $X$ converging to $x$. We want to show that $f(\mathbb{F}):=\{f(F)\mid F\in\mathbb{F}\}$ 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 $f^{{-1}}(U)\in\mathbb{F}$ by assumption. This means that $ff^{{-1}}(U)\in f(\mathbb{F})$. Since $ff^{{-1}}(U)\subseteq U\subseteq N$, we see that $N\in f(\mathbb{F})$ 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 $\mathbb{F}$ be the neighborhood filter of $x$. So $\mathbb{F}\to x$. By assumption, $f(\mathbb{F})\to f(x)$. Since $V$ is an open neighborhood of $f(x)$, we have $V\in f(\mathbb{F})$, or $f(F)\subseteq V$ for some $F\in\mathbb{F}$. 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. ∎

## Mathematics Subject Classification

26A15*no label found*54C05

*no label found*

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