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sequentially compact metric spaces are totally bounded

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

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

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Statement

Proposition

Using excluded middle and dependent choice then:

Let (X,d)(X,d) be a metric space which is sequentially compact. Then it is totally bounded metric space.

Proof

Assume that (X,d)(X,d) were not totally bounded. This would mean that there existed a positive real number ϵ>0\epsilon \gt 0 such that for every finite subset SXS \subset X we had that XX is not the union of the open balls of radius ϵ\epsilon around the elements of this finite subset

XsSB x (ϵ). X \neq \underset{s \in S}{\cup} B^\circ_x(\epsilon) \,.

This would mean that we could inductively choose elements

  • x 0Xx_0 \in X

  • x 1X\B x 0 (ϵ)x_1 \in X\backslash B^\circ_{x_0}(\epsilon)

  • x n+1X\(i=0nB x i (ϵ))x_{n+1} \in X \backslash \left( \underoverset{i = 0}{n}{\cup} B^\circ_{x_i}(\epsilon)\right)

This constituted a sequence (x i)(x_i), and by assumption of sequential compactness this would have a converging sub-sequence. But this would then be a contradiction, since by the above assumption that XX is not totally bounded we have d(x j,x k)>ϵd(x_j, x_k) \gt \epsilon for all jkj \neq k. Hence we have a proof by contradiction that (X,d)(X,d) in fact is totally bounded.

Revised on April 14, 2017 12:45:14 by Urs Schreiber (46.183.103.8)