Fruits of procrastination

Today we will discuss compactness in the metric setting. Why metric? Because metric spaces lend themselves more easily to visualisation than other spaces.

Let us imagine a metric space X with points scattered all over it. If we can find an infinite number of such points and construct disjoint open sets centred on them, then X cannot be compact.

Hence what does it mean to be compact in a metric setting?

Compactness implies that an infinite number of points can’t be ‘far’ away from each other. There can only be a finite number of “clumps” of points such that each neighbourhood, however small, contains an infinite number of such “clumped-together” points. So should you peer at one cump through a microscope, however, strongly you magnify the clump, you will not see discrete points. You will see an impossibly dense patch that shall remain a solid continuus clump of points. 

Complete metric spaces are Baire spaces- a discussion of the proof.

I refer to the proof of the statement “Every complete metric space is a Baire space.”
The proof of this statement, as given in “Introduction to Banach spaces and their Geometry”, by Bernard Beauzamy, is

Let U_1,U_2,\dots be a countable set of _open_ dense subsets of complete metric space X. Take any open set W. We will prove that (\bigcap U_i)\cap W is non-empty.U_1\cap W\neq\emptyset. For point x_1\in U_1\cap W, construct, \overline{B}(x_1,r_1)\subset U_1\cap W. Here \overline{B}(x_1,r_1) is the closure of B(x_1,r_1). Not take the intersection of U_2 and \overline{B}(x_1,r_1), and construct \overline{B}(x_2,r_2)\subset U_2\cap \overline{B}(x_1,r_1). Similarly, for every n\in\Bbb{N}, construct \overline{B}(x_n,r_n)\subset U_n\cap \overline{B}(x_{n-1},r_{n-1}). Also, ensure that 0<r_k<\frac{1}{k}; in other words ensure that r_k converges to 0.

It is clear that the points \{x_1,x_2,\dots\} form a cauchy sequence. As X is complete, the limit l of this point exists in X. This point is present in all U_n, and also in all \overline{B}(x_n,r_n) (as a result, it is present in W). Hence, it is present in \bigcap U_i. Hence (\bigcap U_i)\cap W is not empty. This proves the theorem.

 

We have taken one open set W, and proved, (\bigcap U_i)\cap W is not empty. Should we take another open set T, we will again be able to prove (\bigcap U_i)\cap T is not empty. Note that this does not in any way imply that (\bigcap U_i)\cap W=(\bigcap U_i)\cap T. All that it says is that any open set W has a non-empty intersection with (\bigcap U_i), which is sufficient for (\bigcap U_i) to be a dense set in X.

Why do we not take B(x_k,r_k) instead of \overline{B}(x_k,r_k)? This is because if the balls are not closed, then it would be difficult to prove the limit point of the sequence \{x_1,x_2,\dots\} is also present in \bigcap_i B(x_i,r_i). Try this for yourself. Also, it is ok to take closed balls because if U_i has a non-empty intersection with B(x_{i-1},r_{i-1}), then it definitely has a non-empty intersection with \overline{B}(x_{i-1},r_{i-1}), and we are only concerned with getting an intersection so that we can construct another closed ball inside it. It is elementary to prove that the limit point of \{x_1,x_2,\dots\} has to lie inside the intersection of all the closed balls (hint: \overline{B}(x_k,r_k) contains all x_j of the cauchy sequence \{x_1,x_2,x_3,\dots\}, where j\geq k).