The Baire Category Theorem

Baire Category Theorem

Let $X$ be a completed metric space, let $\{U_n{\}}_{u=1}^{\infty }$ be a countable family of open and dense sets in $X$. Then ${\cap }_{n=1}^{\infty }{U}_n$ is also dense in X, and in particular non-empty. (Here we can consider the definition of dense as $\forall y\in X,\delta >0$, such that $B(y,\delta )\cap U\ne \varnothing$)

Remark

1. The proof of Baire Category Theorem is based on the fact that for $X$ a complete metric space, $\{F_n{\}}_{n=1}^{\infty }$ is a decreasing family of closed sets in $X$ ${\cap }_{n=1}^{\infty }{F}_n\ne \varnothing$
2. Completeness assumption here in Baire Category Theorem can sometimes be weaken.

Nowhere Dense

$X$ be a metric space, $E\subset X$, we say $E$ is nowhere dense if $\overline{E}$ has no interior point.

First Category and Second Category

$E$ is said to be of first category (or meager) if $E$ can be written as a countable union of nowhere dense sets, $E$ is said to be of second category if it’s not of the first category. $E$ is said to be generic if $X\setminus E$ is of first category

Equivalent Version of Baire Category Theorem

A complete metric space can not be of the first category. Equivalently, for $X$ a complete metric space, $X=U_{n=1}^{\infty }{F}_n$ for $F_n$ closed, then there exits $F_n$ that it contains an interior point.

Proof Assume $F_n$ contains no interior point for all $n$, consider $G_n=X/F_n$, $G_n$ open and dense so ${\cap }_{n=1}^{\infty }{G}_n=X/{\cup }_{n=1}^{\infty }{F}_n=\varnothing$, from Baire category, we get a contradiction.

Proposition

1. If $E$ is generic subset of a complete metric space $X$, then $E$ is of the second category. In other words, if $X/E$ is of the first category, then $E$ is of the second category. ($X/E$ is a countable union of nowhere dense sets so if $E$ is of the first category it’s also a countable union of nowhere dense sets, then $X$ is a countable union of nowhere dense set, contradtion.)
2. If a set $E$ in a complete metric space $X$ is a countable intersection of open dense sets, then $E$ is of the second category. ($E={\cap }_{n=1}^{\infty }{U}_n$ for $U_n$ open and dense, then $E$ dense in X, $\overline{E}=X$, which means $X/E$ has no interior point, so of the first category, then $E$ of the second category.)
3. There exists $E$ that $E$ is of the second category but $E$ is not generic. (Consider the example $X=[0,1]\cup [2,3]$, $E=[0,1]$ is complete thus of the second category, but $X/E=[2,3]$ also of the second category)

Corollary

${\mathbb{R}}^d$ is uncountable. Any complete metric space without isolated points is uncountable.

Proof If it’s countable, then ${\mathbb{R}}^d={\cup }_{n=1}^{\infty }{F}_n$ with $F_n$ the closed set only contains one point, by previous version of Baire Category, we get contradiction.

Corollary

1. There exists continuous nowhere differentiable function on $\mathbb{R}$.
2. Let $\{f_n\}$ be a sequence of continuous function on$[0,1]$, suppose $\{f_n\}$ converges pointwise to $f$ at every point in $[0,1]$, then $f$ may not be continuous at every $x\in [0,1]$ but $f$ is continuous at a generic point in $[0,1]$.

Helinger-Toeplitz Theorem

Let $A$ be an everywhere defined linear operator on a Hilbert space \newcommand{\H}{\mathcal{H}}\H such that $⟨x,Ay⟩=⟨Ax,y⟩$ holds for all x,y\in\H. Then $A$ is bounded.

Proof By the closed graph theorem, it suffices to show $\Gamma (A)$ is closed in \H\times\H. Suppose $\{(x_n,A{x}_n){\}}_{n=1}^{\infty }$ is a sequence in $\Gamma (A)$ that converges to $(x,y)$. Then $x_n\to x$ and $A{x}_n\to y$. By the continuity of the inner product, we have $⟨A(x_n-x),A(x_n-x)⟩=⟨A^2(x_n-x),x_n-x⟩\to 0.$Therefore by triangle inequality, we have $\Vert y-Ax\Vert \le \Vert y-A{x}_n\Vert +\Vert A{x}_n-Ax\Vert \to 0.$This shows that $y=Ax$, hence $(x,y)\in \Gamma (A)$, which implies that $\Gamma (A)$ is closed. $\square$

Basis for Banach Space

A Hamel basis for a vector space $X$ is a collection $\mathcal{H}$ of vectors in $X$, such that any $x\in X$ can be written as a unique finite linear combination of elements in $\mathcal{H}$. A Banach space cannot have a countable Hamel basis.