Sequences as Continuous Functions: A Novel Proof That Closed Sets Contain Their Limits

This essay will use the results in my previous blogs : "Beyond Sequences: A Topological Approach to Density Arguments" and "Proving Homeomorphism with Yoneda Lemma: The Unification of epsilon-delta and epsilon-N Formulation". to give an approach for the really basic fact:

Proposition.

Let $F\subseteq X$$F\subseteq X$ be a closed set, $(x_n{)}_{n\in \mathbb{N}}$$(x_n)_{n\in\mathbb N}$ be a convergence sequence in $X$$X$ with $\mathrm{\forall }n,\in \mathbb{N},x_n\in F$$\forall n,\in\mathbb N,x_n\in F$. then we have $\begin{array}{c}{x}_0=\underset{n\to \mathrm{\infty }}{lim}{x}_n\in F\end{array}$$x_0=\lim_{n\to\infty}x_n\in F\\$.

Proof. Notice a convergence sequence is just a continuous function from ${\mathbb{N}}^{\ast }:=\{0\}\cup \{1/n:n\in \mathbb{N}\}$$\mathbb N^*:=\{0\}\cup\{1/n:n\in\mathbb N\}$ with $f(1/n)=x_n,f(0)=x_0$$f(1/n)=x_n,f(0)=x_0$.

Then we have $f^{-1}(F)$$f^{-1}(F)$ is closed as well and $\{1/n:n\in \mathbb{N}\}\subseteq f^{-1}(F)$$\{1/n:n\in\mathbb N\}\subseteq f^{-1}(F)$. Hence the closure of $\{1/n:n\in \mathbb{N}\}={\mathbb{N}}^{\ast }\subseteq f^{-1}(F)◻$$\{1/n:n\in\mathbb N\}=\mathbb N^*\subseteq f^{-1}(F)\quad\square$