nLab closed intervals are compact topological spaces

Theorem

For any $a \lt b \in \mathbb{R}$ the closed interval

$$[a,b] \subset \mathbb{R}$$

regarded with its subspace topology of Euclidean space with its metric topology is a compact topological space.

Proof

Since all the closed intervals are homeomorphic it is sufficient to show the statement for $[0,1]$. Hence let $\{U_i \subset [0,1]\}_{i \in I}$ be an open cover. We need to show that it has an open subcover.

Say that an element $x \in [0,1]$ is admissible if the closed sub-interval $[0,x]$ is covered by finitely many of the $U_i$. In this terminology, what we need to show is that $1$ is admissible.

Observe from the definition that

1. 0 is admissible,

2. if $y \lt x \in [0,1]$ and $x$ is admissible, then also $y$ is admissible.

This means that the set of admissible $x$ forms either

1. an open interval $[0,g)$

2. or a closed interval $[0,g]$,

for some $g \in [0,1]$. We need to show that the latter is true, and for $g = 1$. We do so by observing that the alternatives lead to contradictions:

1. Assume that the set of admissible values were an open interval $[0,g)$. Pick an $i_0 \in I$ such that $g \in U_{i_0}$ (this exists because of the covering property). Since such $U_{i_0}$ is an open neighbourhood of $g$, there is a positive real number $\epsilon$ such that the open ball $B^\circ_g(\epsilon) \subset U_{i_0}$ is still contained in the patch. It follows that there is an element $x \in B^\circ_g(\epsilon) \cap [0,g) \subset U_{i_0} \cap [0,g)$ and such that there is a finite subset $J \subset I$ with $\{U_i \subset [0,1]\}_{i \in J \subset I}$ a finite open cover of $[0,x)$. It follows that $\{U_i \subset [0,1]\}_{i \in J \subset I} \sqcup \{U_{i_0}\}$ were a finite open cover of $[0,g]$, hence that $g$ itself were still admissible, in contradiction to the assumption.

2. Assume that the set of admissible values were a closed interval $[0,g]$ for $g \lt 1$. By assumption there would then be a finite set $J \subset I$ such that $\{U_i \subset [0,1]\}_{i \in J \subset I}$ were a finite cover of $[0,g]$. Hence there would be an index $i_g \in J$ such that $g \in U_{i_g}$. But then by the nature of open subsets in the Euclidean space $\mathbb{R}$, this $U_{i_g}$ would also contain an open ball $B^\circ_g(\epsilon) = (g-\epsilon, g + \epsilon)$. This would mean that the set of admissible values includes the open interval $[0,g+ \epsilon)$, contradicting the assumption.

This gives a proof by contradiction.