Continuous Functions Between Topological Spaces

# Continuous Functions Between Topological Spaces

## Continuous Functions

Definition: Let $E$ and $F$ be topological spaces. A function $f : E \to V$ is Continuous at $x$ if for every neighbourhood $V_{f(x)}$ of $f(x)$ in $F$ there exists a neighbourhood $U_x$ of $x$ in $E$ such that $f(u) \in V$ for all $u \in U$, or equivalently, $f(U_x) \subseteq V_{f(x)}$. The function $f$ is said to be Continuous on $E$ if it is continuous at every point $x \in E$. |

Proposition 1: Let $E$ and $F$ be topological spaces and let $f : E \to F$. Then $f$ is continuous on $E$ if and only if for every open set $V$ in $F$, the preimage, $f^{-1}(V)$, is an open set in $E$. |

**Proof:**$\Rightarrow$: Let $V$ be an open set in $F$. We want to show that $f^{-1}(V)$ is open in $E$. So let $x \in f^{-1}(V)$. Since $f$ is continuous on $E$, it is continuous at $x$ in particular, so for the neighbourhood $V$ of $f(x)$ there exists an open neighbourhood $U_x$ of $x$ in $E$ such that $f(U_x) \subseteq V$. So:

\begin{align} \quad f^{-1}(V) = \bigcup_{x \in f^{-1}(V)} U_x \end{align}

- Which shows that $f^{-1}(V)$ is an arbitrary union of open sets in $E$, so that $f^{-1}(V)$ is open in $E$.

- $\Leftarrow$ Let $x \in E$ and let $V_{f(x)} \subseteq F$ be a neighbourhood of $f(x)$. Then there exists an open set $V'_{f(x)}$ in $F$ such that $x \in V'_{f(x)} \subseteq V_{f(x)}$, and by assumption, $f^{-1}(V'_{f(x)})$ is an open set in $E$ which contains $x$. So set $U_x := f^{-1}(V'_{f(x)})$. Then $U_x$ is an (open) neighbourhood of $x$ with $f(U_x) \subseteq V'_{f(x)} \subseteq V_{f(x)}$. So $f$ is continuous at $x$ and is thus continuous on all of $E$. $\blacksquare$

Proposition 2: Let $E$ and $F$ be topological spaces and let $f : E \to F$. Then $f$ is continuous on $E$ if and only if for every closed set $V$ in $F$, the preimage, $f^{-1}(V)$, is a closed set in $E$. |