# Exercise 3.2.13 Introduction to Real Analysis by Jiri Lebl

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.

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.