Weak Konig's Lemma and Sequential Compactness

Question

May I ask how is "Sequential Compactness" related to "Weal Konig's Lemma"? If we have a bounded set in real number space, how do we prove that every infinite sequence in this set has an infinite subsequence such that this subsequence converges to some point in the set via weak Konig's lemma, which states that if a binary tree has arbitrarily long full finite paths, then it as an infinite path?

Answer 1

Let $\langle x_n:n\in\omega\rangle$ be a bounded sequence in $\Bbb R$, and let $X=\{x_n:n\in\omega\}$. The result is trivial if $X$ is finite, so assume that $X$ is infinite. $X$ is bounded, there is a closed interval $I_{\langle\rangle}$ such that $X\subseteq I_{\langle\rangle}$. Given an interval $I_\sigma$ for some finite sequence $\sigma$ of zeroes and ones, bisect $I_\sigma$, let $I_{\sigma^\frown 0}$ be the left half, and let $I_{\sigma^\frown 1}$ be the right half. Let

$$\mathscr{I}=\big\{I_\sigma:\sigma\in{^{<\omega}\{0,1\}}\big\}\,;$$

$\langle\mathscr{I},\supseteq\rangle$ is a binary tree of height $\omega$. Let $\mathscr{I}_0=\{I_\sigma\in\mathscr{I}:I\cap X\text{ is infinite}\}$. It’s straightforward to verify that $\mathscr{I}_0$ is a subtree of $\mathscr{I}$ with arbitrarily long finite branches, so it has an infinite branch, from which a convergent subsequence is easily constructed.