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Would I be right in stating that for first-countable spaces when finding the limit of a function it can equivalently be reduced to finding a limit involving sequences:

For example if looking for the limit of the function $$\lim\limits_{x \rightarrow a}f(x) = \lim\limits_{x \rightarrow a} x^{2}\sin(\frac{1}{x})$$ it is equivalent to consider finding the the limit of the sequence $$f(x_{n}) = (x_{n})^{2}\sin(\frac{1}{x_{n}})$$

for a sequence $x_{n} \rightarrow a$ (both of which will be 0).

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  • $\begingroup$ I think hyphenating "first-countable" is a good idea. "The first-countable space that we are considering" doesn't mean "the first space that is countable among those that we are considering". ${}\qquad{}$ $\endgroup$ Commented Oct 2, 2014 at 17:10

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You want to show that $\lim_{x \to a} f(x) = L$ iff for any sequence $x_n \to a$, you have $\lim_{n} f(x_n) = L$.

The $\Rightarrow$ direction is straightforward.

For the $\Leftarrow$ direction, use the contrapositive.

Then there exists some $\epsilon>0$ such that for any neighbourhood $U$ of $a$, we have some $x \in U$ such that $|f(x)-L| \ge \epsilon$.

Now construct a sequence $x_n \to a$ using a local base: Let $B_n$ be a countable base at $a$ and let $U_n = B_1 \cap \cdots \cap B_n$, which is open. If we choose any $x_n \in U_n$, then it is straightforward to see that $x_n \to a$.

Now choose $x_n \in U_n$ to be the element such that $|f(x_n)-L| \ge \epsilon$.

Hence there exists some $\epsilon>0$ and a sequence $x_n \to a$ such that such that $|f(x_n) - L| \ge \epsilon$ for all $n$.

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  • $\begingroup$ Thanks for your response. Would I be right in stating that when considering the $\Leftarrow$ direction, that $a$ is usually considered an adherence point of the first-countable space (therefore $a$ is in the closure of the first-countable space). Therefore there would always exists a sequence such that $x_{n} \rightarrow a$, so this implication never holds vacuously? $\endgroup$
    – Alex
    Commented Oct 2, 2014 at 18:48
  • $\begingroup$ Not sure how to answer your question. If $a$ is isolated, then it just says that $f(a) = L$. $\endgroup$
    – copper.hat
    Commented Oct 2, 2014 at 18:56
  • $\begingroup$ Are you not sure my question makes sense or are you not sure of the answer? Since it is first countable and if we assume it is an adherence point then we should be able to construct a sequence $x_{n} \rightarrow a$ such that $x_{n} \neq a$? $\endgroup$
    – Alex
    Commented Oct 2, 2014 at 19:03
  • $\begingroup$ Well, an adherence point only makes sense in the context of some subset of the space (the closure of a space makes no sense unless there is come 'containing' topological space), so I don't know what you mean there. There always exists a sequence $x_n \to a$, specifically $a,a,a,a....$. $\endgroup$
    – copper.hat
    Commented Oct 2, 2014 at 20:27
  • $\begingroup$ I am referring to some subspace of a first-countable space. I think it can be shown that there always exists a sequence $x_{n} \rightarrow a$ such that $x_{n} \neq a$ for any adherence point, even if it is an isolated point. If you see the wiki entry for 'first-countable spaces'. $\endgroup$
    – Alex
    Commented Oct 2, 2014 at 20:41

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