Question

A metric space (X, d) is totally bounded if, given ε>0, there exists a finite subset = $F_{\varepsilon } =\left \{x_{1}, x_{2}, ..., x_{n}\right \}$ of X, called an ε-net, such that for each x∈X there exists $x_{i}$ ∈$F_{\varepsilon }$ such that d(x,$x_{i}$) < ε. Prove that if Y is a subset of a totally bounded space X then, given ε>0, the subset Y has an ε-net and therefore Y is also totally bounded.

Answer 1

1 - x is a totally bounded space. Hence, for every f70 x has E-net. Y EX claim Y is totally bounded. let Eso he arbitrary. X is totally bounded, there enist a finite set Fe = {x1, M2-xa} of X st. for each XeX 7 ni E FE such that olla, nikt 1.

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