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.

  • $\begingroup$ Missin $f(x)$ in the limit. $\endgroup$ – leo May 9 at 5:41

The first part of your proof is fine.

As for the converse, try proving its contrapositive.

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  • $\begingroup$ Thanks for the suggestion! I've edited my original proof to accommodate your suggestion. Is the "converse" direction of this proof correct or is it missing some details? $\endgroup$ – Ricky_Nelson May 6 at 20:52

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