Totally bounded set

**jackie** · March 9th 2011, 10:56 PM

I try to prove that every subset of a totally bounded set is totally bounded, but I am not really sure with my argument here. Could someone help me check it?

Suppose A is a subset of a totally bounded set X. Since X is totally bounded, there exists an $\frac{\epsilon}{2}$ -net $F=\{x_1,x_2,..,x_n\}$ that covers X, i.e $X\subset \bigcup \{B(x_i,\epsilon/2):x_i\in F\}$ . Since A is a subset of X, A is also a subset of the union above. So, $A\cap B(x_i,\epsilon/2)$ for some $x_i \in F$ . Let $y\in A\cap B(x_i,\epsilon/2)$ . Then $B(y,\epsilon)$ covers A since this ball contains every ball with radius $\epsilon/2$ above. Something does not sound right here since I am not sure if the new ball with radius $\epsilon$ contains all the balls with radius $\epsilon/2$

**FernandoRevilla** · March 9th 2011, 11:42 PM

Removing all balls which do not contain points of $A$ and rearranging the subindices if necessary: $A\subset\bigcup_{i=1}^{m}B(x_i,\epsilon/2)\;(m\leq n)$ .

For all $i=1,\ldots ,m$ there exists one point $z_i\in A\cap B(x_i,\epsilon/2)$ . Then, $B(x_i,\epsilon/2)\subset B(z_i,\epsilon)$ .

In fact, if $y\in B(x_i,\epsilon/2)$ we have:

$d(y,z_i)\leq d(y,x_i)+d(x_i,z_i)<\epsilon/2+\epsilon/2=\epsilon$

So, $A\subset \bigcup_{i=1}^{m}B(x_i,\epsilon/2)\subset \bigcup_{i=1}^{m}B(z_i,\epsilon)$ hence, $A$ is totally bounded.

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