# Equivalence of Definitions of Continuous Mapping between Topological Spaces

## Contents

## Continuity at a Point

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ 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 $\map f x$ in $T_2$, there exists a neighborhood $M$ of $x$ in $T_1$ such that $f \sqbrk M \subseteq N$.

### 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 $f \left({x}\right)$.

## Continuity Everywhere

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 $S_2$.

The following definitions of the concept of **everywhere continuous mapping between topological spaces** are equivalent:

### Definition by Pointwise Continuity

The mapping $f$ is **continuous everywhere** (or simply **continuous**) if and only if $f$ is continuous at every point $x \in S_1$.

### Definition by Open Sets

The mapping $f$ is **continuous on $S_1$** if and only if:

- $U \in \tau_2 \implies f^{-1} \sqbrk U \in \tau_1$

where $f^{-1} \sqbrk U$ denotes the preimage of $U$ under $f$.