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

since the interior is equal to the closure of the complement.

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:

  1. Equivalence of definitions of Baire space
  2. Baire property is open subspace-closed

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