1
$\begingroup$

Let $f:S \to \mathbb{R}$ be a function and $c \in S$, such that for every sequence $\{x_n\}$ in $S$ with $\lim x_n = c$, the sequence $\{f(x_n)\}$ converges. Show that $f$ is continuous at $c$.

Suppose that $f$ is not continuous at $c$. Then, $\exists \epsilon > 0$ s.t. for every $\delta> 0$, $\exists x$ s.t. $|x-c| < \delta$, but $|f(x) - f(c)| \ge \epsilon$. Let $\{x_n\}$ be a sequence s.t. $|x_n - c| < \frac1n$. Then, $|f(x_n) - f(c) | \ge \epsilon$ when $n$ is large enough. But, I think that $f(x_n)$ can still converges to somewhere else other than $f(c)$. How can I proceed from here?

I appreciate if you give some help.

$\endgroup$
2
$\begingroup$

Consider any $(x_{n})$ such that $x_{n}\rightarrow c$, consider also the constant sequence $(c,c,...)$ and consider further that $(y_{n}):=(x_{1},c,x_{2},c,...)$, the later also converges to $c$ and $(f(y_{n}))$ has the constant subsequence $(f(c),f(c),...)$, so $f(y_{n})\rightarrow f(c)$ and also that $f(x_{n})\rightarrow f(c)$ as $(f(x_{1}),f(x_{2}),...)$ is also a subsequence of the convergent sequence $(f(y_{n}))$.

So we have proved that, for any $(x_{n})$ such that $x_{n}\rightarrow c$, then $f(x_{n})\rightarrow f(c)$, this is another characterization of continuity.

$\endgroup$

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.