# Prove a metric space is compact, if every infinite subset in it has a limit point.

This is an exercise in W. Rudin's book. Actually my question is, what is an open cover of a metric space? Since an open set is embedded in a certain metric space, how can it cover those points which lie on the bound of the metric space? Should we define a "larger" metric space?

I have read about compactness of metric spaces on Wikipedia. It says:

A metric space M is compact if every sequence in M has a subsequence that converges to a point in M. This is known as sequential compactness and, in metric spaces (but not in general topological spaces), is equivalent to the topological notions of countable compactness and compactness defined via open covers.

And I'm still confused about it.

Now for a concrete example. Let's say your space is $[0,1]$. An open cover could be $B_{0.5}(0),B_{0.1}(0.5),B_{0.5}(1)$ (which are $[0,0.5[$,$]0.4,0.6[$ and $]0.5,1]$ respectively). Obviously, the ball centre 0 and 1 are "missing" the left side and the right side respectively, but as far as the space is concerned, the left and the right never exist at all: the interval $[0,1]$ is just floating in space, not attached or be part of any $\mathbb{R}$.
The definition of an open cover is a collection of open sets that have union the whole space. It is useful to consider the two metric spaces $\Bbb R$ and $[0,1]$, both with the usual (metric) topology. In $\Bbb R, [0,\frac 12)$ is not open, as there is no neighborhood of $0$ included. In $[0,1], [0,\frac 14)$ is an open set that includes $0$, so $[0,\frac 12)$ is open. In $[0,1]$ the ball of radius $0.2$ around $0.1$ is $[0,0.3)$. On the left side the points $(-0.1,0)$ are not in the space, so we don't know they exist. No, we should not define a larger metric space-we are supposed to do everything of interest in the space we are working in.