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 $f(x_n)$?
Best Answer
Suppose we have two logical statements $A$ and $B$.
Proof by contradiction: $$\neg A \implies \bot$$
Proof by contrapositive: $$ \neg A \implies \neg B$$
The proof you gave was done by contradiction.
Original statement $A$:
for every $\epsilon > 0$, there exists a $\delta(\epsilon) > 0$...
Negation of $A$:
there exists $\epsilon >0$, such that for every $\delta(\epsilon)>0$...
In other words, if $A$ was true, then for every $\epsilon$, it was possible to find $\delta(\epsilon)$ such that the subsequent conditions held. The negation of $A$ claims that there is at least one case where the subsequent implications do not hold, meaning we can find a black sheep $\epsilon$ such that every $\delta(\epsilon)$ will not lead us to the desired implication.