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$\mathbf{\text{Definition}\,\,4.3.6\,\,}$

Let $A$ be a subset of a metric space $X$.

We say that $A$ is totally bounded if for every $\varepsilon\gt0$, we can find a finite number of points $x_i,1\le i\le n$, such that $A\subset\cup_{i=1}^{n}B(x_i,\varepsilon)$.

Clearly, this is a back-door entry of compactness! Do you see why?

We say that a subset $A\subset X$ is an $\varepsilon$-net if $d_A(x)\lt\varepsilon$ for any $x\in X$.

Thus $X$ is totally bounded iff there exists a finite $\varepsilon$-net for every $\varepsilon\gt0.$

where $d_A(x)=\inf\{d(x,a):a\in A\}$

Now my question involves the last line:

Thus $X$ is ... for every $\epsilon>0$.

What does finite $\epsilon$-net stand for? Is it finite number of $\epsilon$-net or $\epsilon$-net containing finite number of elements?

Community
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Shantipriya
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  • The text says that a $\varepsilon$-net is a type of subset of $X$, so a "finite $\varepsilon$-net" is a finite set. – Malice Vidrine Nov 04 '13 at 15:26
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    It is an $\varepsilon$-net with finite number of elements. An $\varepsilon$-net is just a set (that satisfies the additional requirement from the definition), so when you add the word finite to it, you mean that the set contains a finite number of elements. – Dan Shved Nov 04 '13 at 15:26

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"$A\subset X$ is a finite $\varepsilon$-net" means that $A$ is finite and is an $\varepsilon$-net. The finite set depends on $\varepsilon$ in this context.

Davide Giraudo
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