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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\title{Notes on Baire's Category Theory}
\date{November 14, 2020 (revised on \today)}
\author{
Z. Zhang
\thanks{
Department of Applied Mathematics, The Hong Kong Polytechnic
University, Hong Kong, China ({\tt zaikun.zhang@polyu.edu.hk}).
}
}

\begin{document}

\maketitle

\section{Basics}

\begin{definition}
  \label{def:intcl}
  Let~$X$ be a topological space and~$S$ be a set in~$X$.
  \begin{enumerate}
    \item The interior of~$S$ is defined as~$\inter(S) = \bigcup\{U \mathrel{:} U\subset S \text{ and } U \text{ is open in } X\}$.
    \item The closure of~$S$ is defined as~$\cl(S) = \bigcap\{G \mathrel{:} G\supset S \text{ and } G \text{ is closed in } X\}$.
  \end{enumerate}
\end{definition}


\begin{proposition}
  \label{prop:intcl}
  Let~$A$ and~$B$ be two sets in a topological space~$X$.
  \begin{enumerate}
    \item Duality between interior and closure: $\inter(A) = \cl(A^\co)^\co$, $\cl(A)=\inter(A^\co)^\co$.
    \item For any~$x\in X$, $x\in \inter(A)$ if and only if there exists an open set~$U$ such that~$x\in U \subset A$.
    \item For any~$x\in X$, $x\in \cl(A)$ if and only if $U\cap A \neq \emptyset$ for any open set~$U$ such that~$x\in U$.
    \item $\inter(A\cap B) = \inter(A)\cap \inter(B)$.
      %, $\inter(A\cup B)\supset \inter(A)\cup \inter(B)$.
      %; if~$A$ is closed and~$\inter(B) = \emptyset$, then~$\inter(A\cup B) = \inter(A)$.
    \item $\cl(A\cup B) = \cl(A)\cup \cl(B)$.
      %, $\cl(A\cap B) \subset \cl(A)\cap \cl(B)$.
      %; if~$A$ is open and~$\cl(B) = X$, then~$\cl(A\cap B) = \cl(A)$.
    \end{enumerate}
\end{proposition}

\begin{proposition}
  \label{prop:subintcl}
  Let~$X$ be a topological space and~$Y$ be its subspace.
  \begin{enumerate}
    \item For any~$S\subset Y$, $\inter_Y(S) \supset \inter_X(S)$;
      if~$Y$ is open in~$X$, then $\inter_Y(S)=\inter_X(S)$.
    \item For any~$S\subset Y$, $\cl_Y(S) = \cl_X(S) \cap Y$.
  \end{enumerate}
\end{proposition}

In Proposition~\ref{prop:subintcl}, we use a subscript to indicate the topological space which the
interior or closure is defined with respect to. We will always do this if the context does not
suffice to avoid confusion.

\begin{proof}
  1. $\inter_X(S)$ is open in~$X$ and~$\inter_X(S)\subset S\subset Y$. Thus~$\inter_X(S)$ is an open
     subset of~$S$ in~$Y$, leading to~$\inter_X(S)\subset \inter_Y(S)$. If~$Y$ is open,
     then~$\inter_Y(S)$ is an open subset of~$S$ in~$X$, implying that~$\inter_Y(S)\subset
     \inter_X(S)$ and hence~$\inter_Y(S) = \inter_X(S)$.

  2. %First,~$\cl_X(S)\cap Y$ is a closed set in~$Y$ and it contains~$S$.
     Since $\cl_Y(S)$ is closed in~$Y$, there exists a closed set~$G$ in~$X$ such
     that~$\cl_Y(S)= G\cap Y$. Thus~$S\subset  G$, implying that~$\cl_X(S) \subset G$.
     Therefore,~$\cl_X(S)\cap Y \subset G\cap Y = \cl_Y(S)$.  On the other hand,~$\cl_X(S)\cap Y$ is
     a closed set in~$Y$ and it contains~$S$, implying that~$\cl_X(S)\cap Y \supset \cl_Y(S)$.
     Hence~$\cl_X(S)\cap Y = \cl_Y(S)$.
\end{proof}

\begin{definition}
  \label{def:dense}
  %Let~$A$ and~$B$ be two sets in a topological space~$X$ such that~$A\subset B$. If~$\cl(A) \supset B$, then~$A$ is said to be dense in~$B$.
  Let~$A$ and~$B$ be two sets in a topological space~$X$
  \begin{enumerate}
      \item If~$\cl(A) = X$ (equivalently, $\inter(A^\co) = \emptyset$), then~$A$ is said to be dense in~$X$.
    \item If~$A\subset B$ and~$A$ is dense in~$B$ with respect to the subspace topology on $B$,
    then we will simply say that~$A$ is dense in~$B$.
  \end{enumerate}
\end{definition}

\begin{proposition}
  \label{prop:densein}
  Let~$A$ and~$B$ be two sets in a topological space~$X$ such that~$A\subset B$. Then~$A$ is dense
  in~$B$ if and only if~$\cl(A) \supset B$.
\end{proposition}

\begin{proof}
  $A$ is dense in~$B$ $\Leftrightarrow $ $\cl_B(A) = B$ $\Leftrightarrow$ $\cl(A)\cap B = B$
  $\Leftrightarrow$ $\cl(A) \supset B$.
\end{proof}

\begin{proposition}
  \label{prop:dense}
  Let~$A$ and~$B$ be two sets in a topological space~$X$.
  \begin{enumerate}
    \item $A\subset \cl(B)$ if and only if~$\inter(B^\co ) \subset \inter(A^\co)$.
    \item \label{it:abu} $A \subset \cl(B)$ if and only if~$U\cap B \neq \emptyset$ for any open
      set~$U$ such that~$U \cap  A \neq \emptyset$
      %(\ie, any neighborhood of any point in~$A$ has a nonempty intersection with~$B$).
    \item \label{it:open} If~$A$ is open, then $A\subset \cl(B)$ if and only if $U\cap B \neq \emptyset$ for any
    nonempty open set~$U\subset A$.
  \item \label{it:abuu} If~$A\subset \cl(B)$, then $A\cap U \subset \cl(B\cap U)$ for any open set~$U$.
   %\item If~$A$ is dense in~$B$, then~$A\cap U$ is dense in~$B\cap U$ for any open set~$U$.
   \item \label{it:dense}If $A$ is open and~$A\subset \cl(B)$, then~$A\cap B$ is dense in~$A$.
   %  , \ie, $A \subset \cl(A\cap B)$.
   \item \label{it:intcl} $\inter(\cl(A))\cap A$ is dense in~$\inter(\cl(A))$.
  \end{enumerate}
\end{proposition}

\begin{proof}
1. $A\subset \cl(B)$ $\Leftrightarrow$ $(\cl(B))^\co \subset A^\co $
$\Leftrightarrow$ $\inter(B^\co)  \subset A^\co$
$\Leftrightarrow$ $\inter(B^\co)  \subset \inter(A^\co)$.

2. Suppose that~$A\subset \cl(B)$. For any open set~$U$ such that~$U\cap A\neq \emptyset$, we
   have~$U\cap \cl(B)\neq \emptyset$, and hence~$U\cap B\neq \emptyset$.
   If~$A\not\subset \cl(B)$, then~$U=(\cl(B))^\co$ is an open set such that~$U\cap A \neq \emptyset$
   while~$U\cap B = \emptyset$.

3. Suppose that~$A\subset \cl(B)$. For any nonempty open set~$U\subset A$, we have~$U\cap A = U\neq \emptyset$,
and hence~$U\cap B\neq \emptyset$ according to~\ref{it:abu}.
   If~$A\not\subset \cl(B)$, then~$U=A\cap (\cl(B))^\co$ is a nonempty open set such that~$U\subset A$
   while~$U\cap B = \emptyset$.

4. Since~$A \subset \cl(B)$, for any open set~$V$ such that~$A\cap U\cap V\neq \emptyset$, we have
according to~\ref{it:open} that~$B\cap U\cap V\neq \emptyset$. This implies~$A\cap U \subset \cl(B\cap U)$
according to~\ref{it:open}.

5. According to~\ref{it:abuu}, $A = A\cap A \subset \cl(A\cap B)$ since~$A$ is open.

6. Since~$\inter(\cl(A))$ is open, and~$\inter(\cl(A)) \subset \cl(A)$, we know from 5 that
$\inter(\cl(A))\cap A$ is dense in~$\inter(\cl(A))$.
\end{proof}

\section{Cantor's theorem and its consequences}

\begin{theorem}[Cantor's theorem]
  \label{th:cantor}
  Let~$X$ be a topological space, and~$\{C_n\}$ be a sequence of
  nonempty closed sets such that~$C_{n+1}\subset C_n$ for each~$n\ge 1$.
\begin{enumerate}
  \item If~$X$ is complete metric space and~$\diam(C_n)\to 0$, then~$\bigcap_{n=1}^\infty C_n$ is
    a singleton.
  \item If each~$C_n$ is compact, then~$\bigcap_{n=1}^\infty C_n$ is nonempty.
\end{enumerate}
\end{theorem}

\begin{proof}
    1.~For each~$n\ge 1$, pick a point~$x_n\in C_n$. For any integers~$m$ and~$n$ such that~$m\ge
    n \ge 1$, we have~$\{x_m,x_n\}\subset C_m \bigcup C_n = C_n$ and hence
    \[
        \dist(x_m,x_n) \;\le\; \diam(C_n) \;\to\; 0 \quad \text{as }~ n\to\infty.
    \]
    Thus the sequence~$\{x_n\}$ is Cauchy, converging to a point~$x\in X$ due to the
    completeness of~$X$. For each~$n\ge 1$, since~$C_n$ is closed
    and~$\{x_k\}_{k\ge n}\subset \bigcup_{k\ge n} X_k = X_n$, we know
    that~$x\in C_n$. Thus~$x\in \bigcap_{n=1}^\infty C_n$. Noting that~$\diam(\bigcap_{n=1}
    ^\infty C_n) \le\diam(C_n) \to 0$, we have~$\bigcap_{n=1}^\infty C_n = \{x\}$.

    2.~For each~$n\ge 1$, pick a point~$x_n\in C_n$. Then~$\{x_n\} \subset X_1$. Due to the
    compactness of~$X_1$, there exists a subsequence~$\{x_{i_k}\}$ of~$\{x_n\}$ that converges to a
    point~$x\in X$. For each~$n\ge 1$, since~$C_n$ is closed and~$\{x_{i_k}\}_{k\ge n}\subset C_{i_n}
    \subset C_n$, we know that~$x\in C_n$ and hence~$x\in \bigcap_{n=1}^\infty C_n$.
    Thus~$\bigcap_{n=1}^\infty C_n$ is nonempty.
\end{proof}

\begin{theorem}[{\cite[Theorems~1.4.5--1.4.6]{Zalinescu_2002}}]
  \label{th:intclcm}
  Let~$X$ be a complete metric space and $\{S_n\}_{n=1}^\infty$ be a sequence of sets in~$X$.
  \begin{enumerate}
      \item\label{it:intcm} If each~$S_n$ is open, then $\cl(\bigcap_{n=1}^\infty S_n)$ and~$\bigcap_{n=1}^\infty \cl(S_n)$ have the same interior.
%      that is $\inter (\cl(\bigcap_{n=1}^\infty S_n)) = \inter(\bigcap_{n=1}^\infty \cl(S_n))$.
    \item If each~$S_n$ is closed, then $\inter(\bigcup_{n=1}^\infty S_n)$ and~$\bigcup_{n=1}^\infty \inter(S_n)$ have the same closure.
%      that is  $\cl (\inter(\bigcup_{n=1}^\infty S_n)) = \cl(\bigcup_{n=1}^\infty \inter(S_n))$.
  \end{enumerate}
\end{theorem}

\begin{proof}
    Due to the duality between closure and interior, we only prove~\ref{it:intcm}, for which it suffices to show that
  $\inter(\bigcap_{n=1}^\infty \cl(S_n))\subset \cl(\bigcap_{n=1}^\infty S_n)$.
  By item~\ref{it:open} of Proposition~\ref{prop:dense}, we only need to prove for a given
  nonempty open set~$U \subset \bigcap_{n=1}^\infty \cl(S_n)$ that%
  \begin{equation}
    \label{eq:usnonempty}
    U\mcap\left(\mcap_{n=1}^\infty S_n\right) \neq \emptyset.
  \end{equation}
  To this end, we will define a sequence of closed balls~$\{B_n\}_{n=0}^\infty$~such~that%
  \begin{equation}
    \label{eq:nest}
  0< \diam(B_n)< 2^{-n},
  \quad
  B_{n+1} \subset B_{n} \subset  U\mcap\left(\mcap_{k=1}^n S_k\right)
  \quad \text{ for each } \quad  n \ge 0,
  \end{equation}
  where $\bigcap_{k=1}^0 S_k=X$.
  We will obtain~\eqref{eq:usnonempty} once this is done, as
  Cantor's theorem will yield
    \begin{equation}
      \label{eq:nestnonempty}
      \emptyset \;\neq \;\mcap_{n=1}^\infty B_n \;\subset\; U\mcap \left(\mcap_{n=1}^\infty
      S_n\right).
    \end{equation}

    We define $\{B_n\}$ inductively.
    As~$U$ is a nonempty open set, we can take a closed ball~$B_0\subset U$
    such that~$0<\diam(B_0)< 1$.
    Assume that the closed ball~$B_n$ is already defined for an~$n\ge 0$
    so that~$0<\diam(B_n)< 2^{-n}$ and~$B_n\subset U\cap(\bigcap_{k=1}^n S_k)$.
    Since~$U\subset\cl(S_{n+1})$ and~$\inter(B_n)$ is a nonempty open subset of~$U$,
    we have that~$\inter(B_n) \cap S_{n+1} \neq \emptyset$. Noting that~$\inter(B_n)\cap S_{n+1}$ is open, we
    can take a closed ball~$B_{n+1}\subset \inter(B_n)\cap S_{n+1}$ such that~$0<\diam(B_{n+1})< 2^{-(n+1)}$.
    It is easy to see that~$B_{n+1}\subset B_n$ and
    %$B_{n+1}\subset B_{n}\cap S_{n+1} \subset [U\cap (\bigcap_{k=1}^n S_{k})]\cap S_{n+1} = U\cap(\bigcap_{k=1}^{n+1} S_k)$.
    $B_{n+1}\subset B_{n}\cap S_{n+1} \subset U\cap(\bigcap_{k=1}^{n+1} S_k)$.
     This finishes the induction and completes the proof.
\end{proof}

\begin{theorem}
  \label{th:intcllch} Theorem~\ref{th:intclcm} still holds if~$X$ is a locally compact Hausdorff
  space.
\end{theorem}

\begin{proof}
  The proof duplicates that of Theorem~\ref{th:intclcm}, except that~$\{B_n\}$ is now a sequence of
  compact sets with nonempty interior such that
  $B_{n+1} \subset B_n \subset U \cap (\bigcap_{k=1}^n S_k)$, the existence of which can be
 established by the local compactness of~$X$.  Since~$X$ is a Hausdorff space, each~$B_n$ is closed,
 and hence~$\bigcap_{n=1}^\infty B_n$ is nonempty by Cantor's theorem.
\end{proof}

\begin{corollary}
    \label{coro:intcl}
    Let~$X$ be a complete metric space or a locally compact Hausdorff space,
    and~$\{S_n\}_{n=1}^\infty$ be a sequence of sets in~$X$.
    \begin{enumerate}
        \item\label{it:dense} If each~$S_n$ is open and dense, then~$\bigcap_{n=1}^\infty S_n$ is dense.
        \item\label{it:emptyint} If each~$S_n$ is closed and has empty interior, then~$\bigcup_{n=1}^\infty S_n$ has empty interior.
    \end{enumerate}
\end{corollary}

\begin{proof}
    \ref{it:dense}.~According to Theorems~\ref{th:intclcm} and~\ref{th:intcllch},
    $\cl(\bigcap_{n=1}^\infty S_n) \supset \inter( \bigcap_{n=1}^\infty \cl(S_n)) = X$.

    \ref{it:emptyint}.~According to Theorems~\ref{th:intclcm} and~\ref{th:intcllch},
    $\inter(\bigcup_{n=1}^\infty S_n) \subset \cl(\bigcap_{n=1}^\infty \inter(S_n)) = \emptyset$.
    Alternatively, we can use item~\ref{it:dense} and the duality between interior and closure.
\end{proof}


\section{Baire's category theorem}

\begin{definition}
Let~$X$ be a topological space.
\begin{enumerate}
  \item A set in~$X$ is said to be rare (or \emph{nowhere dense}) if its closure has empty interior.
  \item A set in~$X$ is said to be meager if it is a countable union of rare sets.
  \item The complement of a meager set is called a comeager (or \emph{residual}) set.
  \item A meager set is also said to be of the first category; other sets are
    of the second category.
\end{enumerate}
\end{definition}


\begin{proposition}
 Rare sets, subsets of meager sets, and countable unions of meager sets are all meager sets.
\end{proposition}

\begin{example}
  Let~$\{r_n\}_{n=1}^\infty$ be an enumeration of all the rational numbers.
  Define
  \begin{equation*}
    S = \bigcap_{m=1}^\infty \bigcup_{n=1}^\infty \left(r_n - \frac{1}{m\,2^n},\;
    r_n+\frac{1}{m\,2^n}\right).
  \end{equation*}
  It is easy to check that~$S$ is comeager in~$\real$. It is dense in~$\real$ but with zero Lebesgue measure.
\end{example}


\begin{definition}
  A topological space~$X$ is called a Baire space if every meager set in~$X$ has empty interior.
\end{definition}

\begin{proposition}
    \label{prop:equiv}
   Let~$X$ be a topological space. The following statements are equivalent.
  \begin{enumerate}
    \item $X$ is a Baire space.
    \item Any comeager set in~$X$ is dense.
    \item Any nonempty open set in~$X$ is not meager.
    \item Any countable intersection of dense open sets in~$X$ is still dense.
    \item Any countable union of rare closed sets in~$X$ has empty interior.
\end{enumerate}
\end{proposition}

\begin{proof}
  1 $\Rightarrow$ 2.
  Let~$S$ be a comeager set in~$X$. Then~$S^\co$ is a meager
  set in~$X$. Thus~$S$ has empty interior, which implies that~$S$ is dense in~$X$.

  2 $\Rightarrow$ 3.
  Let~$U$ be an open set in~$X$.
  If~$U$ is meager, then~$U^\co$ is dense in~$X$. Thus~$U=\!\inter(U) \!= \emptyset$.

  3 $\Rightarrow$ 4.
  Let~$\{U_n\}$ be a sequence of dense open sets in~$X$. Then~$(\bigcap_{n=1}^\infty U_n)^\co$ is
  meager. Its interior is an open meager set in~$X$, and hence empty. Thus~$\bigcap_{n=1}^\infty
  U_n$  is dense.

  4 $\Rightarrow$ 5.
  Let~$\{C_n\}$ be a sequence of rare closed sets in~$X$. Then~$\{C_n^\co\}$ is a seqeunce
  of dense open sets. Thus~$\bigcap_{n=1}^\infty C_n^\co$ is dense in~$X$. Taking the complement, we
  know that~$\bigcup_{n=1}^\infty C_n$ has empty interior.

  5 $\Rightarrow$ 1.
  Let~$S$ be a meager set in~$X$. Then there exists a sequence of rare sets~$\{S_n\}$ such that
  $S=\bigcup_{n=1}^\infty S_n$. Thus~$S\subset \bigcup_{n=1}^\infty \cl(S_n)$, the latter of which has
  empty interior because~$\{\cl(S_n)\}$ is a sequence of rare closed sets.
  Hence~$S$ has empty interior.
\end{proof}

\begin{proposition}
  \label{prop:intclr}
  Let~$X$ be a topological space and~$S$ be a set in~$X$. Then~$S$ is rare in~$X$ if and only
  if~$\inter(\cl(S)) \cap S = \emptyset$.
\end{proposition}

\begin{proof}
  By item~\ref{it:intcl} of Proposition~\ref{prop:dense}, $\inter(\cl(S)) \cap S$ is dense
  in~$\inter(\cl(S))$. Thus $\inter(\cl(S)) = \emptyset$ if and only if $\inter(\cl(S)) \cap S = \emptyset$.
\end{proof}

\begin{lemma}
  \label{lem:openb}
  Let~$X$ be a topological space, $Y$ be its open subspace, and~$S$ be a subset of~$Y$.
  \begin{enumerate}
    \item \label{it:rare}$S$ is rare in $X$ if and only if~$S$ is rare in~$Y$.
    \item $S$ is meager in $X$ if and only if~$S$ is meager in~$Y$.
  \end{enumerate}
\end{lemma}

\begin{proof}
  1. Since~$Y$ is open in~$X$, we have
 \begin{equation}
   \label{eq:subrare}
     \inter_Y(\cl_Y(S)) = \inter_X(\cl_Y(S)) \;=\; \inter_X(\cl_X(S)\cap Y) = \inter_X(\cl_X(S))\cap Y.
 \end{equation}
   If~$S$ is rare in~$X$, then~$\inter_X(\cl_X(S))=\emptyset$, and hence~$\inter_Y(\cl_Y(S))
   = \emptyset$ by~\eqref{eq:subrare},
   implying that~$S$ is rare in~$Y$.
   If~$S$ is rare in~$Y$, then~\eqref{eq:subrare} leads
   to~$\inter_{X}(\cl_X(S))\cap Y = \emptyset$, which implies~$\inter_X(\cl_X(S))\cap
   S = \emptyset$,  ensuring that~$S$ is rare in~$X$ by Proposition~\ref{prop:intclr}.

   2. If~$S$ is meager in~$X$, then~$S= \bigcup_{n=1}^\infty S_n$ with a sequence of rare
   sets~$\{S_n\}$ in~$X$. Each~$S_n$ is a subset of~$S$, and hence of~$Y$, and it is rare
   in~$Y$ according to~\ref{it:rare}. Thus~$S$ is meager in~$Y$. If~$S$ is meager in~$Y$, then~$S
   = \bigcup_{n=1}^\infty S_n$ with a sequence of rare sets~$\{S_n\}$ in~$Y$. Each~$S_n$ is rare in
   in~$X$ according to~\ref{it:rare}. Thus~$S$ is meager in~$X$.
\end{proof}


\begin{theorem}
  \label{th:intcl}
   Let~$X$ be a topological space. The following statements are equivalent.
   \begin{enumerate}
     \item $X$ is a Baire space.
     \item Any open subspace of~$X$ is a Baire space.
     \item For any sequence~$\{S_n\}$ of open sets in~$X$, $\cl(\bigcap_{n=1}^\infty S_n)$ and
       $\bigcap_{n=1}^\infty \cl(S_n)$ have the same interior.
     \item For any sequence~$\{S_n\}$ of closed sets in~$X$, $\inter(\bigcup_{n=1}^\infty S_n)$ and
       $\bigcup_{n=1}^\infty \inter(S_n)$ have the same closure.
   \end{enumerate}
\end{theorem}

\begin{proof}
  1~$\Rightarrow$~2. Let~$Y$ be an open subspace of~$X$, and~$S$ be a meager set in~$Y$.
  Then~$S$ is meager in~$X$ by Lemma~\ref{lem:openb}. Since~$X$ is a Baire space, we
  have~$\inter_X(S)=\emptyset$, which implies~$\inter_Y(S)=\emptyset$ as~$Y$ is open in~$X$.
  Hence~$Y$ is a Baire space.

  2~$\Rightarrow$~3. It suffices to prove that~$\inter(\bigcap_{n=1}^\infty \cl(S_n))\subset\cl(\bigcap_{n=1}^\infty S_n)$.
  Let~$Y=\inter(\bigcap_{n=1}^\infty \cl(S_n))$. Then~$Y$ is an open subspace of~$X$ and hence a Baire space.
  Define $T_n = S_n\cap Y$ for each~$n\ge 1$. Then each~$T_n$ is open in~$Y$. Since~$Y$ is open
  in~$X$ and~$Y\subset \cl_X(S_n)$, we know from item~\ref{it:dense} of Proposition~\ref{prop:dense}
  that~$T_n$ is dense in~$Y$. Hence~$\bigcap_{n=1}^\infty T_n$
  is dense in~$Y$. Therefore, $Y \subset \cl_X(\bigcap_{n=1}^\infty T_n) \subset
  \cl_X(\bigcap_{n=1}^\infty S_n)$ as desired.

  3~$\Rightarrow$~4. Obvious by the duality between interior and closure.

  4~$\Rightarrow$~1.~Let~$\{S_n\}$ be a sequence of closed sets in~$X$ with empty interior. Then we
    have
  $\cl(\inter(\bigcup_{n=1}^\infty S_n)) = \cl(\bigcup_{n=1}^\infty \inter(S_n)) = \emptyset$.
  Thus~$\inter(\bigcup_{n=1}^\infty S_n) = \emptyset$. Hence~$X$ is a Baire space.
\end{proof}

Theorems~\ref{th:intclcm}, \ref{th:intcllch}, and~\ref{th:intcl} lead us to Baire's Category
Theorem. It can also be obtained by combining Corollary~\ref{coro:intcl} and Proposition~\ref{prop:equiv}.

\begin{theorem}[Baire's Category Theorem, {\cite[Theorem~48.2]{Munkres_2000}}]
  \label{th:Baire}
  Complete metric spaces and locally compact Hausdorff spaces are Baire spaces.
\end{theorem}


\section{Baire's category in topological vector spaces}

\begin{definition}
 A topological vector space $X$ is a vector space over a topological field $\KK$ (most often the
 real or complex numbers with their standard topologies) that is endowed with a topology such that
 the vector addition $(x,y)\mapsto x+y$ and scalar multiplication $(\lambda, x) \mapsto \lambda x$ are continuous functions, where the domains of these
 functions are endowed with product topologies.
\end{definition}

\begin{proposition}
  Suppose that~$X$ be a topological vector space and~$U$ is a neighborhood of~$0$.
  Then~$X=\bigcup_{n=1}^\infty (nU)$.
\end{proposition}

\begin{proof}
  For any~$x\in X$, $n^{-1} x\to 0$ due to the continuity of the scalar multiplication.
  Since~$U$ is a neighborhood of~$0$,
  $n^{-1} x$ falls inside~$U$ when~$n$ is sufficiently large, which implies~$x\in n U$.
\end{proof}

\begin{theorem}[{\cite[Theorem~11.6.7]{Narici_Beckenstein_2010}}]
  If~$X$ is a topological vector space, then~$X$ is a Baire space if and only if it is not meager.
\end{theorem}

\begin{proof}
 If~$X$ is a Baire space, then it is not meager as an open subset of itself.  If~$X$ is not a Baire
 space, then there exists a nonempty open set~$U$ that is meager. Take~$x\in U$ and set~$V=U-x$.
 Then~$V$ is a meager neighborhood of~$0$. Thus~$X=\bigcup_{n=1}^\infty (nV)$ is meager.
\end{proof}

\section{Examples}

The application of Baire's category theorem often takes the following form. Let~$X$ be a Baire
space, and~$\{C_n\}$ is a sequence of closed sets in~$X$.
\begin{enumerate}
    \item If each~$C_n$ has no interior, then~$\bigcup_{n=1}^\infty C_n$ cannot be~$X$; indeed, it
        has no interior, and its complement is dense in~$X$.
    \item If~$\bigcup_{n=1}^\infty C_n$ has nonempty interior~(for example, it equals~$X$), then at least
        one~$C_n$ has nonempty interior.
\end{enumerate}

\begin{proposition}
    \label{prop:banach}
    An infinite dimensional Banach space cannot have a countable Hamel basis.
\end{proposition}

\begin{proof}
    Let~$X$ be an infinite dimensional Banach space. Assume that~$\{e_n\}$ is a countable Hamel basis of~$X$.
    Then
    \begin{equation}
        \label{eq:union}
        X = \bigcup_{n=1}^\infty \Span\{e_k \mathrel{:} 1\le k \le n\}.
    \end{equation}
    Since~$\Span\{e_k \mathrel{:} 1\le k \le n\}$ is a finite dimensional subspace of~$X$, it is
    closed. Since~$X$ is infinite dimensional, its finite dimensional subspaces have empty interior.
    Thus~\eqref{eq:union} is impossible as~$X$ is a Baire space.
\end{proof}

\begin{proposition}
    \label{prop:nodiff}
    Let~$X$ be the set of all real-valued continuous functions on $[0,1]$.
    The set of nowhere differentiable functions on~$[0,1]$ is dense in~$X$ under the uniform norm.
\end{proposition}

\begin{proof}
    Let~$Y$ be the set of all periodic continuous functions on~$\RR$ with period~$2$.
    Define
    \begin{align*}
        R \;=\; & \{f \in Y \mathrel{:} f \text{ is not right differentiable  at any } x \in [0,1]\},
        \\
        L \;=\; & \{f \in Y \mathrel{:} f \text{ is not left differentiable  at any } x \in [0,1]\}.
    \end{align*}
    Then the restriction of any function in~$R\cap L$ to~$[0,1]$ is nowhere differentiable
    on~$[0,1]$.
    Therefore, it suffices to prove that~$R \cap L$ is dense in~$Y$ under the uniform norm.
    To this end, we prove that~$R$ and~$L$ are both residual sets in~$Y$ under the uniform norm.
    (N.B.: The purpose of taking~$Y$, $L$, and~$C$ is to deal with the boundaries~$0$ and~$1$.
    For a function on~$[0,1]$, its differentiability at~$0$ is defined as the right
differentiability at~$0$, and similarly for~$1$.)

    Take~$R$ as an example.
    Define
    \[
        C_n \;=\; \{f \in Y \mathrel{:} \text{there exists an }x\in[0,1] \text{ such that } |f(x+h)
        - f(x)| \le nh \text{ for all } h> 0\}.
    \]
    Then~$R \subset \bigcup_{n=1}^\infty C_n$ (for small~$h$, consider the right differentiability of~$f$; for
    other~$h$, consider the boundedness of~$f$). We only need to that each~$C_n$ is closed and has empty
    interior under the uniform norm.

    Let~$\{f_k\}$ be a sequence in~$C_n$ that converges uniformly to~$f_0$.
    Then there exists a
    sequence~$\{x_k\} \subset[0,1]$ such that
    \begin{equation}
        \label{eq:fn}
      |f_k(x_k+h) - f_k(x_k)| \le nh \quad \text{for all~} h>0.
    \end{equation}
    Assume without loss of generality that~$x_k\to x_0\in[0,1]$. By the uniform
convergence of~$\{f_k\}$ and the continuity of~$f_0$, for all~$h\ge 0$, we have
    \[
    |f_k(x_k + h)- f_0(x_0 + h)| \le |f_k(x_k + h) - f_0(x_k +h)| + |f_0(x_k + h) - f_0(x_0 +h)|
        \to 0 \quad \text{as } k\to \infty.
    \]
    Hence~\eqref{eq:fn} implies that~$|f_0(x_0+h) - f(x_0)| \le nh$ for all~$h>0$. Thus~$f_0\in C_n$,
    and~$C_n$ is closed.

    $C_n$ has no interior, because in any neighborhood of any function~$f\in Y$, there exists a
    continuous piecewise linear function whose slope of each linear piece is larger than~$n$ and
    hence not in~$C_n$. The proof is complete.
\end{proof}


\begin{proposition}[Uniform boundedness priciple]
    \label{prop:ub}
    Let~$\TT$ be a set of bounded linear operators between normed linear spaces~$X$ and~$Y$.
    If~$X$ is complete, and the set~$\{Tx \mathrel{:} T\in \TT\}$ is bounded in~$Y$ for all~$x\in X$,
    then the set~$\TT$ is bounded in the operator norm.
\end{proposition}

\begin{proof}
    Define
    \[
        X_n \;=\; \{x\in X \mathrel{:}  \|Tx\| \le n \text{ for all } T \in \TT\}.
    \]
    Then~$X = \bigcup_{n=1}^\infty X_n$ by assumption.  In addition, each~$X_n$ is closed by the
    continuity of~$T \in \TT$. Since~$X$ is a Baire space, there exists an~$n$ such that~$X_n$
    contains a ball~$\BB(x_0, r)$ with~$r>0$. For any~$T\in\TT$ and any~$x \in X$ with~$\|x\| = 1$, we have
    \[
        \|Tx\| \;=\; r^{-1}\|T(rx)\| \;\le\; r^{-1}\left[\|T(x_0 + rx)\| + \|T(x_0)\|\right] \;\le\; 2r^{-1} n.
    \]
    Hence~$\|T\|\le 2r^{-1}n$ for all~$T\in\TT$.
\end{proof}

\begin{proposition}
    \label{prop:nx}
  Let~$f\mathrel{:} (0,+\infty)\to \real$ be a continuous function. If~$\lim_{n\to \infty}f(nx) = 0$
   for all~$x>0$, then~$\lim_{x\to +\infty} f(x) = 0$.
\end{proposition}

\begin{proof} We give two proofs.

  1. Proof by Cantor's theorem. Assume that~$f(x)\nrightarrow 0$ when~$x\to +\infty$. Then there
     exists a constant~$\epsilon > 0$ such that there exists~$x \ge M$ with~$|f(x)|>\epsilon$
     for any~$M > 0$.
     We will define
     a strictly increasing sequence of integers~$\{n_k\}$ and
     a decreasing nested sequence of closed intervals~$\{[a_k, b_k]\}$
     such that~$b_k>a_k > 0$ and
     \begin{equation}
       |f(n_kx)| > \epsilon \quad \text{for all}\quad x \in [a_k, b_k] \quad \text{and}\quad   k\ge 1.
     \end{equation}
     Once this is done, then there exists~$y\in\bigcap_{k=1}^\infty [a_k, b_k]$, and~$f(n_ky)> \epsilon$ for each~$k\ge 1$, contradicting the assumption.

     We define the aforementioned~$\{[a_k,b_k]\}$ and~$\{n_k\}$ inductively.
     Let~$n_1=1$. By the continuity of~$f$, there exists an interval~$[a_1, b_1]$ such
     that~$b_1>a_1>0$ and~$f(n_1x)> \epsilon$ for all~$x\in [a_1, b_1]$. Assume that~$n_k$
     and~$[a_k, b_k]$ has been defined for a~$k\ge 1$ so that~$b_k>a_k >0$. Note that there exists
     $M>0$ such that
     \begin{equation}
       \label{eq:union}
       [M, +\infty) \subset \mcup_{n> n_k} n[a_k, b_k].
     \end{equation}
     Take $a \ge  M$ such that~$|f(a)|> \epsilon$. By the continuity of~$f$, there exists
     $b>a$ such that~$|f(x)|>\epsilon$ for all~$x\in[a, b]$. According to~\eqref{eq:union},
     there exists $n_{k+1} > n_k$ such that~$[a, b] \subset n_{k+1}[a_k, b_k]$.
     Let~$a_{k+1} = a/n_{k+1}$ and~$b_{k+1} = b/n_{k+1}$. Then~$b_{k+1}>a_{k+1}>0$, $[a_{k+1},
     b_{k+1}]\subset[a_k, b_k]$, and~$f(n_{k+1}x) > \epsilon$ for all~$x\in [a_{k+1}, b_{k+1}]$.
     This finishes the induction and completes the proof.

  2. Proof by Baire's Category Theorem. Let~$\epsilon$ be a positive constant. Since~$f(nx)\to 0$
     for all~$x>0$, we have
  \begin{equation}
    \mcup_{N=1}^\infty \mcap_{n=N}^\infty\left\{ x >0 \mathrel{:} |f(nx)| \le \epsilon \right\}
    \;=\; (0,+\infty).
  \end{equation}
  By the continuity of~$f$, $\bigcap_{n=N}^\infty \{x> 0 \mathrel{:} |f(nx)\le \epsilon|\}$ is closed for
  each~$N \ge 1$.
  As~$\real$ is a Baire space, $(0,+\infty)$ is not meager. Thus there exists~$N$ and an open
  interval~$(a, b)$ such that
  \begin{equation}
    (a, b) \subset \mcap_{n=N}^\infty\left\{ x >0 \mathrel{:} |f(nx)| \le \epsilon \right\}.
  \end{equation}
  Therefore
  \begin{equation}
    \label{eq:abe}
    n(a, b) \subset \left\{ x >0 \mathrel{:} |f(x)| \le \epsilon \right\}
    \quad \text{ for each } \quad n\ge N.
  \end{equation}
  Note that there exists~$M >0$ such that
  \begin{equation}
    \label{eq:abm}
    [M, \infty) \subset \mcup_{n=N}^\infty n(a,b).
  \end{equation}
  Combining~\eqref{eq:abe}--\eqref{eq:abm}, we have
  \begin{equation}
    [M, \infty) \subset \left\{ x >0 \mathrel{:} |f(x)| \le \epsilon \right\},
  \end{equation}
  which completes the proof.

\end{proof}

Proposition~\ref{prop:nx} can be strengthened to the following one, which can also be proved by
Baire's Category Theorem.

\begin{proposition}
  Let~$f\mathrel{:} (0,+\infty)\to \real$ be a continuous function. If~$\lim_{n\to \infty}f(nx)$
  exists for all~$x>0$, then~$\lim_{x\to +\infty} f(x)$ exists.
\end{proposition}


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