This post was inspired by a talk given by Professor David Damanik from Rice University. In this post, we will prove the Baire Category theorem for complete metric spaces and briefly talk about applications.
Notation. Let
be a metric space and
. Then,
denotes the open ball centered at
with radius
.
Definition. A space
is Baire if and only if the countable intersection of dense open sets in
is also dense.
For example, the space
is Baire. Consider two dense open sets
and
. The intersection of the two is
which is also dense in
.
Theorem. Any complete metric space is a Baire Space.
proof. We prove by taking a sequence of balls that form an inclusive chain. Call the space
and let
be a sequence of dense open sets. Recall that a subset is dense if and only if every nonempty open subset intersects with the subset. Therefore, take an arbitrary open set
and show that it intersects with each
at a common point for every
. The two sets
has a nonempty open intersection. Take a closed ball
included in the intersect. Repeat the procedure with the open ball
and
to obtain
, and so on with
and
to obtain
.
Clearly, the sequence of closed balls form an inclusive chain.
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The space is complete, so
for some
as
goes to infinity. By construction,
is common to all of the closed balls, therefore to all the sets
and the initial open set
. □
Lets reformulate the result. A
set denotes a countable intersection of open sets. The Baire category theorem says that if the component sets of the open sets are dense, then the resulting
set is also dense.
The Baire category theorem can be used to study decomposition of operators. A Schrodinger operator decomposes the space into eigenstates that correspond to pure point, singular continuous, absolutely continuous spectra. The Wonderland theorem (by Barry Simon) proves that the eigenstates corresponding to pure point and absolute continuous spectra form a dense subset in the space of all states. The proof involves the Baire category theorem.
References and further reading:
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