# The Heine-Borel Theorem for the real line

Hi everyone I'd like to know if the following argument is correct and also I'm very interested in a constructive approach for (2)$\Rightarrow$(1) (a link or a hint it will sufficient for me) I was thinking for a while but I cannot figure out some way to do it. Thanks in advance.

Definitions: A subset $X$ of the real line is bounded if we have $X\subset [-M, M]$ for some real number $M>0$.

Let $X\subset \mathbb{R}$ and let $x'\in \mathbb{R}$, $x'$ is an adherent point of $X$ iff $\,\forall\varepsilon>0\, ,\exists x\in X \text{ s.t.} \; d(x',x)\le \varepsilon$. We say that $\overline{X}$ is the closure of $X$ if contain all the adherent points of $X$.

Theorem (The Heine-Borel Theorem for the line): Let $X$ be a subset of $\mathbb{R}$. Then the following two statements are equivalent:

(1) $X$ is closed and bounded

(2) Given any sequence $(a_n)$ of real numbers which takes values from $X$ (i.e., $a_n\in X$ for all natural numbers), there exists a subsequence ($a_{n_j}$) which converges to some number $L$ in $X$.

Proof:

(1)$\Rightarrow$(2)

Let $(a_n)_{n=0}^\infty$ be a sequence where $a_n\in X$ for all the natural numbers. Since $X$ is bounded, $|a_n|\le M$ for all $n$, where $M$ is an upper bound for $X$. Then by the Bolzano–Weierstrass theorem there exist a convergent subsequence $(a_{n_j})$. Let $L=\lim_j a_{n_j},$ then it follows that $L$ is an adherent point of $X$ and since is closed by hypothesis, $L$ is in $X$ as desired.

(2)$\Rightarrow$(1)

Suppose for the sake of contradiction that either $X$ is not closed or is unbounded. If $X$ is unbounded, let define $X_n=\{y\in X: |y|>n\}$ (each one is non-empty). Then we can find using the AC a sequence $(x_n)$ such that $x_n \in X_n$ for each positive integer. By hypothesis we know that there is a subsequence $(x_{n_j})$ which converges to some $L$ in $X$. But $|a_n|>L+1$ for all $n\ge L+1$ so, for $j\ge L+1$ we have $|a_{n_j}|>L+1$, a contradiction. Now if $X$ is not closed then it has at least one adherent point $x$ which is not in the set. Since $x$ is adherent there is a sequence $(a_n)$ of terms in $X$ which converges to $x$, i.e., $a_n\rightarrow x$ and $a_n\in X$ for all $n$. Since any subsequence of a convergent sequence converges to the same value, so $x$ must be in $X$, a contradiction. It follows that $X$ must be closed and bounded. $\Box$

• All is correct. Technically you are studying sequential compactness. About the "constructive" part, if my memory serves me, the use of (numerable) choice is required. Feb 1, 2014 at 8:07
• According to Axiom of Choice by Horst Herrlich, in ZF a sequentially compact subset of $\mathbb{R}$ may fail to be bounded or to be closed. The direction (1)->(2) holds in ZF. Feb 1, 2014 at 21:38
• Martín-BlasPérezPinilla: Thanks for your help and MichaelGreinecker: Thanks for the reference :) Feb 1, 2014 at 22:36