> [!definition] > > Let $X$ be a [[Topological Space|topological space]], then the following are equivalent: > 1. If $\seq{A_j}$ are nowhere [[Dense|dense]], then $\bigcup_{j \in \nat}A_j \ne X$. > 2. If $\seq{A_j}$ are [[Closed Set|closed]] and have empty interior, then $\bigcup_{j \in \nat}A_j \ne X$. > 3. If $\seq{A_j}$ are [[Closed Set|closed]] and $\bigcup_{j \in \nat}A_j = X$, then there exists $N \in \nat$ such that $\bigcup_{j = 1}^NA_j$ has non-empty [[Interior|interior]]. > 4. If $\seq{U_j}$ are [[Open Set|open]] dense subsets, then $\bigcap_{j \in \nat}U_j$ is also dense in $X$. > > If the above holds, then $X$ is **of the second category**, and is known as a **Baire Space**. > > *Proof*. $(1) \Rightarrow (2)$: Since $A_j$ have empty interior, they are nowhere dense. Therefore $\bigcup_{j \in \nat}A_j \ne X$. > > $(2) \Rightarrow (3)$: If $\bigcup_{j \in \nat}A_j = X$, then there must exist $n \in \nat$ such that $A_n$ has non-empty interior. Thus $\bigcup_{j = 1}^n A_j$ has non-empty interior as well. > > $(3) \Rightarrow (4)$: Let $A_j = U_j^c$, then $A_j$ is closed and has empty interior. If $\bigcap_{j \in \nat}U_j$ is not dense in $X$, then there exists $N \in \nat$ such that $\paren{\bigcap_{j = 1}^N U_j}^c$ has non-empty interior, which contradicts the fact that $\seq{U_j}$ are dense. > > $(4) \Rightarrow (1)$: Let $U_j = (\overline{A_j})^c$, then $U_j$ is open and dense. Since $\bigcap_{j \in \nat}U_j$ is dense in $X$, $\bigcap_{j \in \nat}A_j \ne X$. > [!theorem] > > Let $X$ be a Baire space and $U \subset X$ be open, then $U$ is also a Baire space. > > *Proof*. Let $\seq{V_j}$ be open dense subsets of $U$, then $V_j \cup \overline{U}^c$ is open and dense in $X$. Since $X$ is Baire, $\bigcap_{j \in \nat}V_j \cup \overline{U}^c$ is open and dense in $X$ as well. Thus $\bigcap_{j \in \nat}V_j$ is dense in $U$. > [!theoremb] Baire Category Theorem > > Let $X$ be a topological space. Suppose that for every $U \subset X$ open and family $\seq{U_j}$ of open dense subsets, there exists $\seq{V_j}$ of open sets such that > 1. $\overline{V_j} \subset U_j \cap V_{j - 1} \subset U$ for all $j > 1$, with $\overline{V_1} \subset U$. > 2. $\bigcap_{j \in \nat}\overline{V_j}$ is non-empty. > > then $X$ is a Baire space. > > *Proof*. $(1)$ and $(2)$ implies that $U \cap \bigcap_{j \in \nat}\overline{U_j} \ne \emptyset$ for all $U \subset X$ open. > [!theoremb] Baire Category Theorem > > Let $X$ be a locally compact Hausdorff space or a [[Complete Metric Space|completely metrisable]] space, then is a Baire space. > > *Proof*. If $X$ is completely metrisable, fix a metric $d$. > > Let $V_0 = U$. For each $j \in \nat$, since $U_j$ is dense, $U_j \cap V_{j - 1} \ne \emptyset$. Let $x \in U_j \cap V_{j - 1}$, then since $X$ is [[Regular Space|regular]], there exists $V_j \subset U_j \cap V_{j - 1}$ such that $x \in V_j \subset \overline{V_j} \subset U_j \cap V_{j - 1}$. If $X$ is LCH, choose $V_j$ to be [[Precompact|precompact]]. If $X$ is completely metrisable, choose $V_j = B(x, r)$ where $r < 2^{-j}$. Thus $(1)$ of the previous lemma holds. > > If $X$ is LCH, then since $V_1$ is compact and $\seq{V_j}$ satisfies the finite intersection property, $(2)$ holds. > > If $X$ is a complete metric space, then for each $j \in \nat$, there exists $x_j \in V_j$. By our choices, $\seq{x_j}$ is a [[Cauchy Sequence|Cauchy sequence]]. Therefore there exists $x \in \bigcap_{j \in \nat} \overline{V_j}$ and $(2)$ holds. > [!theoremb] Theorem > > Let $X$ be a Baire space, $(Y, d)$ be a [[Metric Space|metric space]], and $\seq{f_n} \subset C(X, Y)$ such that $f_n \to f \in Y^X$ [[Pointwise Convergence|pointwise]], then there exists $E \subset X$ dense such that $f$ is [[Continuity|continuous]] at every $x \in E$. > > *Proof*. Let $\eps > 0$, $N \in \nat$, and define > $ > A_N(\eps) = \bracs{x \in X: d(f_m(x), f_n(x)) \le \eps \forall m, n \ge N} > $ > By [[Continuity|continuity]] of $\seq{f_n}$ each $A_N(\eps)$ is an intersection of closed sets, and is closed. > > For each $\eps > 0$, since each $\seq{f_n(x)}$ is Cauchy, $X = \bigcup_{N \in \nat}A_n(\eps)$. Let $U \subset X$ be open, then $\bigcup_{N \in \nat}A_N(\eps) \cap U = U$. As $U$ is Baire, there exists $N \in \nat$ such that $[A_N(\eps)]^o \cap U \ne \emptyset$. > > Let $U(\eps) = \bigcup_{N \in \nat}[A_N(\eps)]^o$, then $U(\eps)$ is open and intersects every open set in $X$. Thus $U(\eps)$ is dense. > > Let $C = \bigcap_{k \in \nat}U(1/k)$ and $x \in C$, then for each $k \in \nat$, there exists $N \in \nat$ such that $x \in [A_N(1/k)]^o$. By continuity of $f_N$, there exists $V \in \cn^o(x)$ such that $d(f_N(x), f_N(y)) \le 1/k$ for all $y \in V$. In addition, for every $y \in A_N(1/k)$, $d(f_N(y), f(y)) \le 1/k$. By the triangle inequality, for every $y \in [A_N(1/k)]^o \cap V \in \cn^o(x)$, $d(f(x), f(y)) \le 3/k$. Therefore $f$ is continuous at $x$. > > As $X$ is Baire, $C$ is dense.