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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
    
Yes. his proof seems written in a strange way, with the "for every "\ldelta" there is an x with 0 < |x-a| < \delta$, since we do not exactly have an $x$, but only the $x_k$. The idea of building a sequence is correct though. –  Calvin Lin Dec 30 '12 at 20:08

2 Answers 2

up vote 1 down vote accepted

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.

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Very helpful! thanks –  Hawk Dec 30 '12 at 20:16

Definition of continuity:

If $f:X\subseteq \mathbb{R}\to \mathbb{R}$ and $a\in X$ then $f$ is continuous at $a$ if $$\forall \epsilon>0\ \exists \delta>0\ :\forall x\in X\ \left|x-a\right|<\delta\implies \left|f(x)-f(a)\right|<\epsilon$$

The negation: If $f:X\subseteq \mathbb{R}\to \mathbb{R}$ and $a\in X$ then $f$ is discontinuous at $a$ if $$\exists\epsilon>0:\ \forall \delta>0\ \exists x\in X:\ \left|x-a\right|<\delta\text{ and } \left|f(x)-f(a)\right|\ge\epsilon$$

So $\epsilon$ can be thought as constant (independent of $\delta$), $\delta$ is arbitrary (changing) and $x=x_{\epsilon,\delta}$

To your last question, yes. If you pick $\delta=1,\frac12,...,\frac1n,...$ you have a decreasing $\delta$ that gives $x_n\to a$ and serves as your contradiction.

Notes: I don't know why mr. Spivak writes $0<\left|x-a\right|$. Most definitions of continuity don't contain this (not that it changes too much).

Second, the argument: "for each $n$ there would be a number $x_n$ such that..." is a bit dark. To make it rigorous consider $\delta=1$, $\delta=\frac12$ etc. and construct such a sequence $x_n$

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