Theorem 4.9.2. Let $X$ be a topological space, $Y$ be a regular space, $A \subset X$, and $f \in C(A; Y)$, then the following are equivalent:
There exists $F \in C(X; Y)$ such that $F|_{A} = f$.
For each $x \in X$, $\lim_{y \to x, y \in A}f(y)$ exists.
Proof. $(\Rightarrow)$: Let $x \in X$ and $\fF = \bracs{U \cap A| U \in \cn(x)}$, then $f(\fF) = F(\fF)$. By continuity of $F$, $F(\fF)$ converges to $F(x)$, so $\lim_{y \to x, x \in A}f(x)$ exists.
$(\Leftarrow)$: Let $F(x) = \lim_{y \to x, y \in A}f(y)$, then $F$ is well-defined and $F|_{A} = f$ by (4) of Definition 4.8.1.
Let $x \in X$ and $V \in \cn(F(x))$. Using (2) of Definition 4.9.1, assume without loss of generality that $V$ is closed. By assumption, there exists $U \in \cn^{o}(x)$ such that $f(U \cap A) \subset V$. In which case, Lemma 4.4.3 implies that $U \in \cn^{o}(y)$ for all $y \in U$. Since every limit point of a filter is a cluster point,
as $V$ is closed. Therefore $F(U) \subset V$, and $F$ is continuous.$\square$