I am trying to develop a non-topological proof for the Sequential Criterion for Functional Limits. Can someone please check the accuracy of my proof? Thanks!
Sequential Criterion for Functional Limits: Let $f: A\to\mathbb{R}$. Given $c$ is a cluster point in $A$. Prove that the following statements are equivalent:
(a) $\lim_{x \rightarrow c} = L$.
(b) For all sequences $(x_n) \subseteq A$ with $x_n \neq c$ and $(x_n)\rightarrow c$ $\implies$ $f(x_n) \rightarrow L$
Proof. Suppose $\lim_{x \to c} = L$. Then, $\forall \epsilon > 0, \exists \delta > 0,$ s.t.$0 < \mid x-c \mid < \delta \implies \mid f(x)-L \mid < \epsilon$. Now, let $(x_n) \subseteq A$ with $x_n \neq c$ and $(x_n)\rightarrow c$. Pick $\delta = \epsilon$. Then, $\forall \epsilon >0, \exists N \in \mathbb{N}$ s.t. $n \geq N \implies 0< \mid x_n - c \mid < \delta$ which implies $\mid f(x_n) - L \mid < \epsilon.$
(Contrapositive) Conversely, suppose $\lim_{x \to c} \neq L$. Then, $\exists \epsilon_0 > 0$ s.t. $ \forall \delta > 0$ it is that $0 < \left| x-c \right| < \delta \implies \left| f(x) - L \right| \geq \epsilon_0$. Note that this proposition is true for all $x \in$ dom $f$. Now, let $(x_n) \subseteq A$ be arbitrary with $x_n \neq c$. Then, based on our argument so far, we are justified in concluding that $\forall n$, it is that $0< \left| x_n - c \right| < \delta \implies \left| f(x_n) - L \right| \geq \epsilon_0$ are we are done.
