# Ngô Quốc Anh

## November 4, 2014

### Baire properties for open subspaces

Filed under: Giải Tích 1 — Tags: — Ngô Quốc Anh @ 7:53

This post deals with a classical problem in functional analysis: The Baire space. I am not going to reproduce what we can learn and read from wikipedia; however, to make the post self-contained, following is what the Baire space is.

Loosely speaking, a Baire space $X$ is a topological space in which any one of the following three equivalent conditions is satisfied:

1. Whenever the union of countably many closed subsets of $X$ has an interior point, then one of the closed subsets must have an interior point, i.e. if

$\displaystyle \text{int}\Big(\bigcup_{n \geqslant 1} C_n \Big) \ne \emptyset,$

then $\text{int}(C_n) \ne \emptyset$ for some $n$. Here by $C$ we mean a closed subset in $X$.

2. The union of every countable collection of closed sets with empty interior has empty interior, that is to say, i.e if $\text{int}(C_n) = \emptyset$ for all $n$, then

$\displaystyle \text{int}\Big(\bigcup_{n \geqslant 1} C_n \Big) =\emptyset.$

3. Every intersection of countably many dense open sets is dense, i.e.

$\displaystyle \overline{\bigcap_{n \geqslant 1} O_n} = X$

provided $\overline{O_n}= X$ for every $n$. Here by $O$ we mean an open subset in $X$.

What I am going to do is to show that every open subset of a Baire space is itself a Baire space, of course, under the subspace topology inherited from $X$. Hence, at the very beginning, we assume throughout this topic that $X$ is a Baire space; hence admits all three equivalent conditions above.

First, given an open subset $O \subset X$, we show that:

Result 1. If $(C_n)_n$ is a sequence of closed subsets in $O$ (not necessarily close in $X$) such that

$\displaystyle \text{int}\Big(\bigcup_{n \geqslant 1} C_n \Big) \ne \emptyset,$

then $\text{int}(C_n) \ne \emptyset$ for some $n$.

A proof for this result is based on contradiction; hence, it suffices to prove the following second result:

Result 2. If $(C_n)_n$ is a sequence of closed subsets in $O$ (not necessarily close in $X$) such that $\text{int}(C_n)= \emptyset$ for all $n$, then

$\displaystyle \text{int}\Big(\bigcup_{n \geqslant 1} C_n \Big)= \emptyset.$

A proof for this result is basically simple which is mainly based on the third result below. For the sake of clarity, we divide the proof into several steps:

Step 2.1. We first denote $O_n = O \backslash C_n$ which is open in $O$. Observe that the set $O_n$ is dense in $O$. Indeed,

$\displaystyle \overline{O_n} = \overline{O \backslash C_n}=O \backslash \text{int}(C_n) = O$

Step 2.2. We now consider the intersection $\bigcap_{n \geqslant 1} O_n$ in $O$. In view of the third result below, there holds

$\displaystyle \overline{\bigcap_{n \geqslant 1} O_n} = O,$

hence by De Morgan’s laws

$\displaystyle \overline{O \backslash \bigcap_{n \geqslant 1} C_n} = O,$

which is again equivalent to saying that

$\displaystyle O \Big\backslash \text{int}\Big(\bigcup_{n \geqslant 1} C_n \Big) = O,$

thus

$\displaystyle\text{int}\Big(\bigcup_{n \geqslant 1} C_n \Big) = \emptyset$

as we wish.

It is important to note that in all two results above, the interior action is taken with respect to the sub-topology. However, it is not difficult to show that the interior of a set in the sub-topology generated by an open set remains the same for the full-topology, i.e. in our context

$\displaystyle \text{int}_X(A) = \text{int}_O(A).$

As can be seen, our proof above relies on the fact that any intersection of countable many open dense subsets remains dense and this is the content of the third result right below.

We now state the third result:

Result 3. If $(O_n)_n$ is a sequence of open subsets in $O$ such that $\overline{C_n} = X$ for all $n$, then

$\displaystyle \overline{\bigcap_{n \geqslant 1} O_n} = O.$

A proof for this result is rather elegant which, for the sake of clarity, is divided into several steps as follows:

Step 3.1. We again denote $O_X = X \backslash \overline O$ which is open in $X$. Notice that each $O_n$ is open in $O$ and since $O$ is also open in $X$, we know that $O_n$ is open in $X$. This together with the fact that $O_X$ is open in $X$ conclude that $O_n \cup O_X$ is open in $X$.

Step 3.2. In this step, we show that the set $O_n \cup O_X$ is in fact dense in $X$. To do show, let $O^X$ be any open subset in $X$, we need to prove that

$\displaystyle ({O_n} \cup {O_X}) \cap {O^X} \ne \emptyset$

There are two possibilities: Either $O_X \cap O^X \ne \emptyset$ or $O^X \subset \overline O$. In the former case, there is nothing to do. In the latter case, $O^X$ cannot be part of the boundary $\partial O$; hence $O^X \cap O = W$ is an non-empty open subset of $O$. Now since $O_n$ is dense in $O$, apparently $O_n \cap W \ne \emptyset$; hence showing that $(O_n \cup O_X) \cap {O^X} \ne \emptyset$ as claimed.

Step 3.3. Now we make use of the fact that $X$ is a Baire space to conclude that

$\displaystyle \overline{\bigcap_{n \geqslant 1} \big( O_n \cup O_X \big) } = X.$

We now use De Morgan’s laws to rewrite $X$ as the following

$\displaystyle X = \overline{ \Big( \bigcap_{n \geqslant 1} O_n \Big) \cup O_X}.$

Step 3.4. In the last step, we prove that

$\displaystyle \overline{\bigcap_{n \geqslant 1} O_n} = O.$

Again, to do so, we let $O_O$ be an open subset in $O$, out aim is to show that $\big(\bigcap_{n \geqslant 1} O_n \big) \cap O_O \ne \emptyset$. Clearly, $O_O$ is open in $X$; hence

$\displaystyle \Big(\Big( \bigcap_{n \geqslant 1} O_n \Big) \cup O_X \Big) \cap O_O \ne \emptyset;$

hence

$\displaystyle \Big(\Big( \bigcap_{n \geqslant 1} O_n \Big) \cap O_O \Big) \cup \Big( O_X \cap O_O \Big) \ne \emptyset.$

Since $O_O\subset O$, there holds $O_X \cap O_O = \emptyset$; hence $\big( \bigcap_{n \geqslant 1} O_n \big) \cap O_O \ne \emptyset$.

Finally, we comment that all three conditions the the Baire space are equivalent. I will provide a proof for this result when I have time.

References: