# Heine-Cantor Theorem: introducing a subsequence in proof.

I need some help with the demonstration of the Heine-Cantor theorem:

Given a continuous function $f: A\rightarrow B$, with $A, B \subseteq \Bbb R$ and $A$ compact (closed and limited set), $f$ is uniformly continuous in $A$.

Proof:

Suppose $f$ is not uniformly continuous. Then:

$\exists \epsilon _{0}>0$ for which $\forall \delta>0, \exists x, y \in A$, with $|x-y| < \delta$ and $|f(x) - f(y)| \ge \epsilon_{0}$

Now choose $\delta \in \{\frac{1}{n}, n \in \Bbb N^{+}$} and build two sequences $(x_{n})_{n}$ and $(y_{n})_{n}$ such that $|x_{n} - y_{n}| < \delta$ and $|f(x_{n}) - f(y_{n})| \ge \epsilon_{0}$ for each $n \in \Bbb N^+$

As seen that $A$ is compact, then we can extract a subsequence of $x_{n}$, that we call $(x_{k_{n}})_{n}$, that converges to a certain point $x_{0} \in A$.

It follows that (as $|x_{n} - y_{n}| < \delta$): $x_{k_n} - \frac{1}{k_n} \le y_{k_n} \le x_{k_n} + \frac{1}{k_n}$ so $(y_{k_n})_n$ converges to $x_0$ too.

Because $f$ is continuous we get that $\lim_{n \to \infty}{f(x_{k_n})} = f(x) = \lim_{n \to \infty}{f(y_{k_n})}$. Therefore: $\lim_{n \to \infty}{[f(x_{k_n}) - f(y_{k_n})]} = 0$ but this result is absurd because of the hypothesis ($|f(x_{n}) - f(y_{n})| \ge \epsilon_{0}$ for each $n \in \Bbb N^+$ ). $\blacksquare$

The part in bold is what I do not understand. Why do we need to extract a subsequence? I mean: if $x_{k_n}$ converges to a value $x_0$ don't any subsequence of $x_n$ and $x_n$ itself do the same? So what's the point in building subsequences? And: why have we built two sequences of points? That's what is not clear to me.

• From what is stated, though, you just have a sequence of points $x_n$. The sequence is not necessarily convergent but it does necessarily have a convergent subsequence. Mar 24 '17 at 21:47

Think about the sequence $$1, {1\over 2}, 1, {1\over 3}, 1, {1\over 4}, 1, {1\over 5}, ...$$ "Half" of it converges to zero: ${1\over 2}, {1\over 3}, {1\over 4}, ...$ But the whole sequence doesn't, because of those $1$s. So while the whole sequence isn't convergent, it has a subsequence which is. And this is why we need to pass to a subsequence in this argument (and in lots of other arguments).

As to your other question: when we assume that $f$ isn't uniformly continuous, that means (roughly) that for each $\delta>0$ we can find points $x, y$ which are "close" but whose $f$-values are "far apart." Note that these $x, y$ depend on $\delta$: if I change $\delta$, I (might) get different $x, y$s. So when we think about a sequence of $\delta$s going to $0$ (the proof above uses ${1\over 2}, {1\over 3}, {1\over 4}, ...$ but we could use any other similar sequence), we immediately get a pair of sequences: the $x$s, and the $y$s, corresponding to each $\delta$. So it's not that we've decided to go looking for a pair of sequences, but rather the assumption of non-uniform continuity gave us a pair of sequences, and so we'll play around with them.