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.
Best Answer
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.