# Every subset $Y$ of a totally bounded metric space $(X,d)$ is also totally bounded

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?

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 $$Y$$.
• They are equivalent. But the way your question is worded I believe you have prove that you get points from $Y$ rather than assume that two notions are equivalent. It is just a question of how the question is interpreted. @Abhay – Kavi Rama Murthy Oct 23 '19 at 10:16