# Proof verification: $(x_n)\rightarrow c$ $\implies$ $f(x_n) \rightarrow L$ iff $\lim_{x \rightarrow c} = L$

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.

• Missin $f(x)$ in the limit.
– leo
May 9, 2020 at 5:41
• Your proof of the converse is wrong. The negation of $\forall x(P(x)\Rightarrow Q(x))$ is not $\forall x(P(x)\Rightarrow\neg Q(x))$ but $\exists x(P(x)\land\neg Q(x)).$ For a correct proof, see e.g. math.stackexchange.com/questions/3504461/… Jan 9 at 19:43

## 1 Answer

The first part of your proof is fine.

As for the converse, try proving its contrapositive.

• 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? May 6, 2020 at 20:52
• The rule on MSE for a proof verification is to type comments, not an answer. Jan 9 at 19:45