0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Missin $f(x)$ in the limit. $\endgroup$ – leo May 9 at 5:41
0
$\begingroup$

The first part of your proof is fine.

As for the converse, try proving its contrapositive.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.