# Alternative Proof of Equivalent definitions of continuity

The proof that we often see for the equivalence of $$\epsilon$$-$$\delta$$ definition of a limit and the definition of a limit in terms of sequences often use proof by contradiction. As an exercise I wanted to try a direct proof.

I'll only prove one direction as this is the part I had a question on. We consider function $$f: \mathbb{R} \rightarrow \mathbb{R}$$ and assume that it is continuous at $$(x,y)$$ in the sequential sense.

Choose any $$\epsilon > 0$$. Further, choose any sequence $$\{x_n\}$$ where $$\lim x_n = x$$ and $$x_n \neq x$$ for all $$n$$. We know that there must exist some $$n_1$$ where $$n > n_1$$ implies $$|f(x_n) - y| < \epsilon$$. Also, there must be some $$n_2$$ for which $$n > n_2$$ implies $$|x_n - x| < \epsilon$$. Now, we let

$$\delta = \sup \{|x_n - x| : n > \max(n_1, n_2)\}.$$

Clearly, $$\delta$$ is finite and larger than zero. Further if we have $$x'$$ such that $$|x' - x| < \delta$$, then there must exist some $$x_{n'}$$ for which $$|x' - x| < |x_{n'} - x|$$.

I want to conclude that $$|f(x') - y| < \epsilon$$, but I'm not sure how to proceed from here.

Suppose that $$f$$ is given by
$$f(x)=\begin{cases} 0,&\text{if }x=\frac1n\text{ for some }n\in\Bbb Z^+\\ 2x,&\text{otherwise;} \end{cases}$$
this is continuous at $$x=0$$. We might happen to choose the sequence $$\left\langle\frac1n:n\in\Bbb Z^+\right\rangle$$, so that $$x_n=\frac1n$$ for each $$n\in\Bbb Z^+$$. Now take $$\epsilon=\frac12$$. We can take $$n_1=1$$ and $$n_2=2$$, so that $$\delta=\frac13$$. Let $$x'=\frac3{10}$$; then $$|x'|<\delta$$, and $$x'<\frac13=x_3$$, so we can take $$n'=3$$. But then
$$|f(x')-f(0)|=\frac35>\frac12=\epsilon\,.$$
Thus, on the basis of what you’ve done so far you cannot hope to prove in general that $$|f(x')-y|<\epsilon$$. I am not at all sure that a direct proof is possible, because somehow you really do need to use the fact that $$f$$ behaves at $$x$$ for all sequences converging to $$x$$.