Every subset $Y$ of a totally bounded metric space $(X,d)$ is also
totally bounded
$X$ is totally bounded $\rightarrow \forall \epsilon > 0 \; \exists n(\epsilon) \in N$ and $\exists x_1\ldots x_n \in X $ such that $$\cup_{i=1}^{n}B_\epsilon(x_i) = X \supseteq Y$$ so $Y$ is totally bounded.
Is this correct?
Best Answer
There are two notions of total boundedness for a subset $Y$ of a metric space $X$. In one definition we just $Y$ is totally bounded and require points from $Y$ such that the $\epsilon$ balls around them cover $Y$. In the other notion we talk about $Y$ being a totally bounded subset of $Y$ (and say $Y$ is totally bounded in $X$) where is no such requirement.
To show that $Y$ with the restriction of the metric $d$ is totally bounded in its own right you have to obtain points from $Y$, but your points $x_i$ may not be in $Y$.
Use what you have done with $\epsilon$ changed to $\epsilon /2$. Without loss of generality assume that $Y$ has at least one point in common with each of the balls $B(x_i,\epsilon /2)$. If $y_i \in B(x_i, \epsilon /2)\cap Y$ show that the balls $B(y_i,\epsilon)$ cover $Y$.