# Analyzing a sequence and continuity proof

I am trying to understand the following proof:

Given $f$ is continuous, prove that for every convergent sequence $(x_n) \to a$ that $\lim_{k\to \infty}f(x_k) = f(a)$

So the prove goes like this (by contradiction and contrapositive?). I found this proof in Spivak's which the whole content will be a quote

If $\lim_{k\to \infty}f(x_k) = f(c)$ were not true, there would be some $\epsilon > 0$ such that for every $\delta$ there is an $x$ with

$0 < |x - a| < \delta$, but $|f(x) - f(a)| > \epsilon$

In particular, for each $n$ there would be a number $x_n$ such that

$0 < |x_n - a| < \delta$, but $|f(x_n) - f(a)| > \epsilon$

Now the sequence $(x_n)$ clearly converges to $c$, but since $|f(x_n) - f(a)| > \epsilon$ for all $n$, the sequence $(f(x_n))$ does not converge to $f(a)$. Which is a contradiction.

When they were doing the contrapositive. They say $\forall \delta >0$, s.t. $\exists \epsilon >0$ etc etc...

Does the epsilon here now depend on $\delta$? If so doesn't that mean it wouldn't make much sense to pick a changing $\delta$? If $\epsilon$ here doesn't change, was the goal just to find some $\delta$ (a decreasing one) as to contradict the non-changing $\epsilon$ ball around $f(x_n)$?

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It is a proof by contradiction. If $\lim_{k \rightarrow \infty} f(x_k) \neq f(c)$, then it means that there exists an $\epsilon$ (which we now fix) such that we can always find a sequence $x_{N_k}$, such that $f(x_{N_k})$ is $\epsilon$ far away from $f(c)$. This in turn shows that $f$ is not continuous. –  Calvin Lin Dec 30 '12 at 19:48
So my original doubt was correct right? –  Hawk Dec 30 '12 at 19:58

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