# Equivalence of Definitions of Continuous Mapping between Topological Spaces/Point

## Contents

## Theorem

Let $T_1 = \left({S_1, \tau_1}\right)$ and $T_2 = \left({S_2, \tau_2}\right)$ be topological spaces.

Let $f: S_1 \to S_2$ be a mapping from $S_1$ to $S_2$.

Let $x \in S_1$.

The following definitions of the concept of **continuity at a point of a topological space** are equivalent:

### Definition using Open Sets

The mapping $f$ is **continuous at (the point) $x$** (with respect to the topologies $\tau_1$ and $\tau_2$) if and only if:

- For every neighborhood $N$ of $f \left({x}\right)$ in $T_2$, there exists a neighborhood $M$ of $x$ in $T_1$ such that $f \left({M}\right) \subseteq N$.

### Definition using Filters

The mapping $f$ is **continuous at (the point) $x$** if and only if for any filter $\mathcal F$ on $T_1$ that converges to $x$, the corresponding image filter $f \left({\mathcal F}\right)$ converges to $f \left({x}\right)$.

Let $X, Y$ be topological spaces.

Let $x \in X$ and let $f: X \to Y$ be a mapping.

This has to be rewritten. In particular: the notation of the proof does not match the notation of the definition pages. The old statement is kept below |

The following are equivalent:

- $(1): \quad f$ is continuous at $x$, i.e. for any neighborhood $U \subseteq Y$ of $f \left({x}\right)$, $f^{-1} \left({U}\right)$ is a neighborhood of $x$.

- $(2): \quad$ For any filter $\mathcal F$ on $X$ that converges to $x$, the corresponding image filter $f \left({\mathcal F}\right)$ converges to $f \left({x}\right)$.

## Proof

### $(1) \implies (2)$

Assume $f$ is continuous at $x$ and that $\mathcal F$ is a filter on $X$ that converges to $x$.

Suppose $U \subseteq Y$ is a neighborhood of $f \left({x}\right)$.

Then, since $f$ is continuous at $x$, $f^{-1} \left({U}\right)$ is a neighborhood of $x$.

Since $\mathcal F$ converges to $x$, this implies that $f^{-1} \left({U}\right) \in \mathcal F$.

By the definition of $f \left({\mathcal F}\right)$ it follows that $U \in f \left({\mathcal F}\right)$.

Thus $f \left({\mathcal F}\right)$ converges to $f \left({x}\right)$.

### $(2) \implies (1)$

The set $\mathcal U_x := \left\{{U \subseteq X: U \text{ is a neighborhood of } x}\right\}$ is a filter on $X$.

We set $\mathcal F := \mathcal U_x$.

By the definition of convergence for filters, $\mathcal F$ converges to $x$.

By assumption this implies that $f \left({\mathcal F}\right)$ converges to $f \left({x}\right)$.

Now let $U \subseteq Y$ a neighborhood of $f \left({x}\right)$.

Then $U \in f \left({\mathcal F}\right)$ and by definition this implies $f^{-1} \left({U}\right) \in \mathcal F = \mathcal U_x$.

Thus $f^{-1} \left({U}\right)$ is a neighborhood of $x$.

Therefore $f$ is continuous at $x$.

$\blacksquare$