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; 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 $j\geq k$).