Baire Category Theorem

In this exercise, we prove the Baire category theorem (i.e. Problem 3).

Let $K$ be a topological space. Recall that a subset $S \subseteq K$ is dense in $K$ if $\bar S = K$ .

Problem 1. Prove that $S$ is dense in $K$ if and only if for any nonempty open set $U \subseteq K$ , $S \cap U \neq \emptyset$ .

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Solution. We prove this statement by contrapositive. For the direction $(\Rightarrow)$ , suppose $\bar S \subsetneq K$ . Then there exists $x \in K \backslash \bar S$ . Since the latter is open, it contains a neighbourhood $U$ of $x$ so that $U \subseteq K \backslash \bar S \subseteq K \backslash S$ , yielding $S \cap U = \emptyset$ , as required.

For the direction $(\Leftarrow)$ , suppose there exists a nonempty open set $U \subseteq K$ such that $S \cap U = \emptyset$ . Then the closed set $K \backslash U$ contains $S$ and thus $\bar S \subseteq K \backslash U$ . Since $U \neq \emptyset$ , there exists $x \in U$ so that $x \notin K \backslash U \supseteq \bar S$ . Hence, $\bar S \subsetneq K \backslash U \subsetneq K$ , as required.

Now suppose $K$ is a complete metric space with metric $d$ .

Problem 2. Prove that if $C \subseteq K$ is closed, then $C$ is complete.

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Solution. Fix any Cauchy sequence $\{x_n\}$ in $C \subseteq K$ . Since $K$ is complete, $x_n \to x$ for some $x \in K$ . Since $C$ is closed and $K$ is a metric space, $x \in C$ . Since every Cauchy sequence in $C$ converges in $C$ , we have that $C$ is complete.

Problem 3. Prove that for any countable collection $\{U_n \}$ of open dense sets in $K$ , $\bigcap_n U_n$ is dense in $K$ .

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Solution. Fix any open set $V \subseteq K$ . Since each $U_k$ is dense, $V \cap U_k \neq \emptyset$ . It suffices to prove that $V \cap \bigcap_n U_n \neq \emptyset$ .

Define $U_0 := V$ . Inductively, since $B(x_k, r_k) \cap U_{k+1} \neq \emptyset$ , find $x_{k+1} \in B(x_k, r_k) \cap U_{k+1}$ and $r_{k+1} \in (0, 2^{-(k+1)})$ such that

$\overline{ B(x_{k+1}, r_{k+1}) } \subseteq B(x_k, r_k) \cap U_{k+1}.$

Inductively, for any $k$ ,

$\displaystyle \overline {B(x_{k+1}, r_{k+1})} \subseteq V \cap \bigcap_{i=1}^k U_i.$

Now consider the sequence $\{x_n\}$ . It is Cauchy since for sufficiently large $m, n$ , $d(x_m, x_n) < 2^{-\min\{m, n\}}$ . Since $\bar V$ is closed, it is complete. Hence, $\{x_n\}$ is Cauchy in $\bar V$ and converges to some $x \in \bar V$ . By a straightforward $\epsilon$ – $K$ verification, $x \in V \cap \bigcap_{n} U_n$ , as required.

Problem 4. Prove that if $K = \bigcup_{i=1}^\infty C_i$ for a countable collection $\{C_i\}$ of closed sets, then at least one $C_i$ has nonempty interior.

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Solution. By assumption, taking complements yields $\bigcap_{i=1}^\infty K \backslash C_i = \emptyset$ . For a contradiction, suppose all $C_i$ have empty interior. We claim that $\overline{ K \backslash C_i } = K$ , so that $K \backslash C_i$ being dense, via Problem 3, yields the desired contradiction $K = \overline{\bigcap_{i=1}^\infty K \backslash C_i} = \bar{\emptyset} = \emptyset$ .

Suppose for a contradiction that $\overline{ K \backslash C_i } \subsetneq K$ . Then there exists $x \in K \backslash \overline{ K \backslash C_i }$ . Since the latter is open, there exists a neighbourhood $U \subseteq K \backslash \overline{ K \backslash C_i }$ of $x$ . Hence, $K \backslash C_i \subseteq \overline{ K \backslash C_i } \subseteq K \backslash U$ implies that $C_i \supseteq U \ni x$ , contradicting the assumption that $C_i$ has empty interior.