closed intervals are compact topological spaces



topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory


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

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

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


Since all the closed intervals are homeomorphic it is sufficient to show the statement for [0,1][0,1]. Hence let {U i[0,1]} iI\{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[0,1]x \in [0,1] is admissible if the closed sub-interval [0,x][0,x] is covered by finitely many of the U iU_i. In this terminology, what we need to show is that 11 is admissible.

Observe from the definition that

  1. 0 is admissible,

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

This means that the set of admissible xx forms either

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

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

for some g[0,1]g \in [0,1]. We need to show that the latter is true, and for g=1g = 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)[0,g). Pick an i 0Ii_0 \in I such that gU i 0g \in U_{i_0} (this exists because of the covering property). Since such U i 0U_{i_0} is an open neighbourhood of gg, there is a positive real number ϵ\epsilon such that the open ball B g (ϵ)U i 0B^\circ_g(\epsilon) \subset U_{i_0} is still contained in the patch. It follows that there is an element xB g (ϵ)[0,g)U i 0[0,g)x \in B^\circ_g(\epsilon) \cap [0,g) \subset U_{i_0} \cap [0,g) and such that there is a finite subset JIJ \subset I with {U i[0,1]} iJI\{U_i \subset [0,1]\}_{i \in J \subset I} a finite open cover of [0,x)[0,x). It follows that {U i[0,1]} iJI{U i 0}\{U_i \subset [0,1]\}_{i \in J \subset I} \sqcup \{U_{i_0}\} were a finite open cover of [0,g][0,g], hence that gg itself were still admissible, in contradiction to the assumption.

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

This gives a proof by contradiction.

Last revised on June 6, 2017 at 04:20:03. See the history of this page for a list of all contributions to it.