Math 2.0

statement proposition

Closed subsets of complete spaces are complete

Statement

Let (X,d)(X, d) be a complete metric space and let FF be a closed subset of XX. Then FF with the subspace metric is complete.

Remarks

Proof: let (xn)(x_n) be a Cauchy sequence in FF. It is Cauchy in XX, so it converges to some x∈Xx \in X. Then x∈Fx \in F, because closed sets contain their limits. For example, [0,1][0, 1], Z\mathbb{Z} and each closed ball of Rn\mathbb{R}^n are complete. The converse is complete subspaces are closed.

Explanations

L1 human

A Cauchy sequence in FF has a limit in the big complete space. Because FF is closed, this limit cannot escape from FF.

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Sources

  1. Web https://en.wikipedia.org/wiki/Complete_metric_space, Some theorems