A Function is Continuous iff it preserves Net Convergence
# Statement
Let $X, Y$ be Topological Spaces and let $f: X \to Y$ be a Function. Then $f$ is a Continuous Function If and Only If for all Nets ${x_{\alpha}}{\alpha \in A} \subset X$, if $\lim\limits{\alpha \in A}x_{\alpha} = x$ then $\lim\limits_{\alpha \in A}f(x_{\alpha}) = f(x)$.
# Proof
First note for Net ${x_{\alpha}}{\alpha \in A} \subset X$, $f \circ x{\bullet}$ is a Net on $Y$ since it is a Function from Directed Partial Ordering $A$ to $Y$.
$(\Rightarrow)$: Suppose $f$ is a Continuous Function. Let ${x_{\alpha}}{\alpha \in A} \subset X$ be a Net. Let $V \subset Y$ be Open so $V \ni f(x)$. Then $f^{-1}(V)$ is Open and $x \in f^{-1}(V)$. Since $x{\alpha}\to x$, there exists $\alpha_{0}$ s.t. for all $\alpha \geq \alpha_{0}$, $x_{\alpha} \in f^{-1}(V)$. Thus for all $\alpha \geq \alpha_{0}$, $f(x_{\alpha}) \in V$. Then, by definition of Net Convergence, $f(x_{\alpha}) \to f(x)$. $\checkmark$
($\Leftarrow$): Let $K \subset Y$ be Closed. Let ${x_{\alpha}}{\alpha \in A} \subset f^{-1}(K)$ be a Net so that $\lim\limits{\alpha \in A} x_{\alpha} = x \in X$. We will show that $x \in f^{-1}(K)$. By assumption, we know $\lim\limits_{\alpha \in A}f(x_{\alpha}) = f(x)$. Since ${x_{\alpha}}{\alpha \in A}$ is a subset of $f^{-1}(K)$, ${f(x{\alpha})}{\alpha \in A} \subset K$. Recall A Set is Closed iff it contains all Net Limits. Thus $f(x) \in K$. This means $x \in f^{-1}(K)$. Since ${x{\alpha}}_{\alpha \in A}$ was arbitrary, then because A Set is Closed iff it contains all Net Limits, $f^{-1}(K)$ is Closed in $X$. Thus, because A Function is Continuous iff it takes Closed Sets back to Closed Sets, $f$ is a Continuous Function. $\checkmark$ $\blacksquare$