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 is a topological space in which any one of the following three equivalent conditions is satisfied:

- Whenever the union of countably many closed subsets of has an interior point, then one of the closed subsets must have an interior point, i.e. if
then for some . Here by we mean a closed subset in .

- The union of every countable collection of closed sets with empty interior has empty interior, that is to say, i.e if for all , then
- Every intersection of countably many dense open sets is dense, i.e.
provided for every . Here by we mean an open subset in .

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 . Hence, at the very beginning, we assume throughout this topic that is a Baire space; hence admits all three equivalent conditions above.

First, given an open subset , we show that:

Result 1. If is a sequence of closed subsets in (not necessarily close in ) such thatthen for some .

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

Result 2. If is a sequence of closed subsets in (not necessarily close in ) such that for all , then

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 which is open in . Observe that the set is dense in . Indeed,

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

*Step 2.2*. We now consider the intersection in . In view of the third result below, there holds

hence by De Morgan’s laws

which is again equivalent to saying that

thus

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

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 is a sequence of open subsets in such that for all , then

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 which is open in . Notice that each is open in and since is also open in , we know that is open in . This together with the fact that is open in conclude that is open in .

*Step 3.2*. In this step, we show that the set is in fact dense in . To do show, let be any open subset in , we need to prove that

There are two possibilities: Either or . In the former case, there is nothing to do. In the latter case, cannot be part of the boundary ; hence is an non-empty open subset of . Now since is dense in , apparently ; hence showing that as claimed.

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

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

*Step 3.4*. In the last step, we prove that

Again, to do so, we let be an open subset in , out aim is to show that . Clearly, is open in ; hence

hence

Since , there holds ; hence .

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**:

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