Definition 4.6.1 (Continuity). Let $X$ and $Y$ be topological spaces, $f: X \to Y$ be a function, and $x \in X$, then the following are equivalent:
For each $V \in \cn(f(x))$, $f^{-1}(V) \in \cn(x)$.
For each filter base $\fB \subset 2^{X}$ converging to $x$, $f(\fB)$ converges to $f(x)$.
If the above holds, then $f$ is continuous at $x \in X$.
The following are also equivalent:
For each $U \subset Y$ open, $f^{-1}(U)$ is open in $X$.
$f$ is continuous at every $x \in X$.
For each convergent filter base $\fB \subset 2^{X}$, $f(\fB)$ is convergent.
If the above holds, then $f$ is continuous.
The collection $C(X; Y)$ is the space of all continuous functions from $X$ to $Y$.
Proof. Local continuity, $(1) \Rightarrow (2)$: Let $V \in \cn(f(x))$, then $f^{-1}(V) \in \cn(x)$. Since $\fB$ converges to $x$, there exists $B \in \fB$ with $B \subset f^{-1}(V)$. In which case, $f(B) \subset V$ and $V \in \fF(\fB)$ by (F1).
Local continuity, $(2) \Rightarrow (1)$: Since $f(\cn(x))$ converges to $f(x)$, for each $U \in \cn(f(x))$, there exists $V \in \cn(x)$ such that $f(V) \subset V$. Thus $V \subset f^{-1}(U) \in \cn(x)$ by (F1).
Global continuity, $(1) \Rightarrow (2)$: Let $U \subset Y$ be open, then $U \in \cn(f(x))$ for all $x \in f^{-1}(U)$. Hence $f^{-1}(U) \in \cn(x)$ for all $x \in f^{-1}(U)$. Thus $U$ is open by Lemma 4.4.3.
Global continuity, $(2) \Rightarrow (1)$: Let $x \in X$ and $V \in \cn(f(x))$, then there exists $U \in \cn(f(x))$ open, so $f^{-1}(V) \supset f^{-1}(U)$ contains an open set that contains $x$. Therefore $f^{-1}(V) \in \cn(x)$.
Global continuity, $(2) \Leftrightarrow (3)$: Follows from local continuity.$\square$