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.
Best Answer
You probably are confused by visualizing the space at some sort of bounded shape, such as a square, and you might have thought that at the boundary of the square, the ball is no longer a ball shape. However, you should remember that a "ball" is just an expression to make it easy to remember what the definition is, rather than any sort of circular ball. Think of each topological space as a world onto itself: then there is no "boundary", and the ball does in fact contains all the points with radius. For example, a metric space might have the discrete metric (distinct element have metric 1), then a ball of radius 0.5 will contains a single point, while a ball of radius 1.5 will contains the whole space. Obviously, the best visualization of such space would be a whole bunch of points scattered around, and there is no circularly shaped ball.
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 $\mathbb{R}$.