[proofplan] We code the Banach-Mazur interval game below a basic open interval as an integer game, so $\mathrm{AD}$ gives a winning strategy for one of the two players. The category analysis of this game says: player I wins exactly when the payoff set is comeagre on some smaller basic interval, while player II wins exactly when the payoff set is meagre in the original interval. Applying this dichotomy to countably many basic intervals, we form the [open set](/page/Open%20Set) $U$ by collecting all basic intervals on which $A$ is comeagre. Countability makes the local meagre errors into one global meagre set, and the maximality of $U$ forces $A$ to be meagre off $U$. [/proofplan]

[step:Fix a countable interval basis and define the local Banach-Mazur games]Let $\mathcal{B}$ denote the countable basis of $\mathbb{R}$ consisting of all nonempty open intervals with rational endpoints. Thus every nonempty open subset of $\mathbb{R}$ contains some element of $\mathcal{B}$. For an interval $L = (a,b) \in \mathcal{B}$, let $|L| := b-a$ denote its Euclidean length. Fix $I \in \mathcal{B}$. Define the Banach-Mazur game $G_I(A)$ as follows. Player I and player II alternately choose intervals, subject to the legality conditions that player I chooses $I_0 \in \mathcal{B}$ with $\overline{I_0} \subset I$ and $|I_0| < 1$, player II chooses $I_1 \in \mathcal{B}$ with $\overline{I_1} \subset I_0$ and $|I_1| < 1/2$, and, continuing inductively, after $I_k$ is chosen, the next player chooses $I_{k+1} \in \mathcal{B}$ with $\overline{I_{k+1}} \subset I_k$ and $|I_{k+1}| < 1/(k+2)$. The closure nesting and the length condition imply that the sequence $(I_k)_{k \in \mathbb{N}}$ determines a unique real number $x \in I$ with \begin{align*} \{x\} = \bigcap_{k=0}^{\infty} \overline{I_k}. \end{align*} Player I wins the play if $x \in A$, and player II wins if $x \notin A$. Because each legal move is chosen from the [countable set](/page/Countable%20Set) $\mathcal{B}$, fix an enumeration $e: \mathbb{N} \to \mathcal{B}$ and code a legal interval move $L \in \mathcal{B}$ by any $n \in \mathbb{N}$ with $e(n)=L$. To obtain a total game on $\mathbb{N}$, declare that the first player who plays a code whose interval is illegal at that position loses immediately; if no illegal code is played, the winner is determined by the resulting real $x$ as above. This total integer game has a payoff subset of $\mathbb{N}^{\mathbb{N}}$, so $\mathrm{AD}$ gives a winning strategy for one of the two players. Restricting such a strategy to legal histories gives a winning strategy in $G_I(A)$. [step:Explain why the coded game is a total game on $\mathbb{N}$][/step]

[guided]We need a game to which $\mathrm{AD}$ applies. The axiom concerns games whose moves are natural numbers, so we choose moves from the countable basis $\mathcal{B}$ of rational open intervals. Fix an interval $I \in \mathcal{B}$. A legal play is a decreasing sequence of basic intervals inside $I$, with closures nested and lengths tending to $0$. The nesting condition $\overline{I_{k+1}} \subset I_k$ guarantees that the closed intervals $\overline{I_k}$ form a decreasing sequence of nonempty compact subsets of $\mathbb{R}$. The length condition $|I_k| \to 0$ forces their intersection to contain at most one point. Since compact nested nonempty closed intervals have nonempty intersection, there is exactly one real number $x \in I$ such that \begin{align*} \{x\} = \bigcap_{k=0}^{\infty} \overline{I_k}. \end{align*} The payoff set for player I is the set of plays whose resulting real $x$ lies in $A$. Since every move is an element of the countable set $\mathcal{B}$, an enumeration of $\mathcal{B}$ converts this into a game on $\mathbb{N}$. Therefore the hypothesis $\mathrm{AD}$ applies and gives a winning strategy for exactly one of the two players.[/guided]

[/step]

[step:Prove the Banach-Mazur category dichotomy for one interval] We use the following local category dichotomy. [claim:Banach-Mazur category dichotomy] Let $I \in \mathcal{B}$ and let $A \subseteq \mathbb{R}$. In the game $G_I(A)$, if player I has a winning strategy, then there exists $J \in \mathcal{B}$ with $\overline{J} \subset I$ such that $J \setminus A$ is meagre in $J$. If player II has a winning strategy, then $A$ is meagre in $I$. [/claim] [proof] First suppose player I has a winning strategy $\sigma$. Let $J \in \mathcal{B}$ be the first interval prescribed by $\sigma$, so $\overline{J} \subset I$. We prove that $J \setminus A$ is meagre in $J$. Let $\mathcal{P}_\sigma$ be the countable set of all finite legal positions compatible with $\sigma$ and ending immediately after a move of player I. For $s \in \mathcal{P}_\sigma$, let $O_s \in \mathcal{B}$ denote the last interval in $s$. Define $R_s \subseteq O_s$ to be the union of all intervals $M \in \mathcal{B}$ for which there exists a legal player-II move $L \in \mathcal{B}$ after $s$ such that $M$ is the next interval prescribed by $\sigma$ after the extended position $(s,L)$. For each $s \in \mathcal{P}_\sigma$, the set $R_s$ is open in $O_s$. It is dense in $O_s$: if $K \in \mathcal{B}$ is nonempty with $\overline{K} \subset O_s$, player II may choose a legal basic subinterval $L \subseteq K$ with sufficiently small length, and then the strategy $\sigma$ prescribes a legal response $M$ with $\overline{M} \subset L \subseteq K$. Hence $K \cap R_s \neq \varnothing$. Therefore $B_s := O_s \setminus R_s$ is nowhere dense in $O_s$ and consequently nowhere dense in $J$, since the boundary of the interval $O_s$ has empty interior in $J$. Define \begin{align*} N_\sigma := \bigcup_{s \in \mathcal{P}_\sigma} B_s. \end{align*} Since $\mathcal{P}_\sigma$ is countable, $N_\sigma$ is meagre in $J$. Let $x \in J \setminus N_\sigma$. Starting with the one-move position $s_0=(J)$, we have $x \in O_{s_0}$ and $x \notin B_{s_0}$, hence $x \in R_{s_0}$. Thus player II can choose a legal move after $s_0$ such that the following $\sigma$-move has last interval containing $x$. Repeating this argument at each later player-I position produces one coherent full legal play following $\sigma$ in which every chosen interval contains $x$. Because the legal length bounds tend to $0$ and the closures are nested, the unique real determined by this play is $x$. Since $\sigma$ is winning for player I, $x \in A$. Hence $J \setminus A \subseteq N_\sigma$, so $J \setminus A$ is meagre in $J$. Now suppose player II has a winning strategy $\tau$. Let $\mathcal{P}_\tau$ be the countable set consisting of the empty initial position, together with all finite legal positions compatible with $\tau$ and ending immediately after a move of player II. For $s \in \mathcal{P}_\tau$, let $O_s := I$ if $s$ is the empty position, and otherwise let $O_s \in \mathcal{B}$ be the last interval in $s$. Define $R_s \subseteq O_s$ to be the union of all intervals $M \in \mathcal{B}$ for which there exists a legal player-I move $L \in \mathcal{B}$ after $s$ such that $M$ is the response interval prescribed by $\tau$ after the extended position $(s,L)$. For each $s \in \mathcal{P}_\tau$, the set $R_s$ is open in $O_s$. It is dense in $O_s$: given a nonempty $K \in \mathcal{B}$ with $\overline{K} \subset O_s$, player I may choose a legal basic subinterval $L \subseteq K$ with sufficiently small length, and then $\tau$ responds with a legal interval $M$ satisfying $\overline{M} \subset L \subseteq K$. Thus $K \cap R_s \neq \varnothing$. Consequently $B_s := O_s \setminus R_s$ is nowhere dense in $O_s$ and nowhere dense in $I$. Define \begin{align*} N_\tau := \bigcup_{s \in \mathcal{P}_\tau} B_s. \end{align*} Since $\mathcal{P}_\tau$ is countable, $N_\tau$ is meagre in $I$. Let $x \in I \setminus N_\tau$. Starting from the empty position, the condition $x \in R_s$ at each current $\tau$-compatible position allows player I to choose a legal move whose following $\tau$-response contains $x$. Iterating gives one coherent full legal play compatible with $\tau$ in which every interval contains $x$. The nesting and length conditions force the real determined by this play to be $x$. Since $\tau$ is winning for player II, $x \notin A$. Hence $A \cap I \subseteq N_\tau$, so $A$ is meagre in $I$. This proves the dichotomy. [/proof] [/step]

[step:Collect all basic intervals on which $A$ is locally comeagre] Define \begin{align*} \mathcal{U} := \{J \in \mathcal{B} : J \setminus A \text{ is meagre in } J\}. \end{align*} Define the open set $U \subseteq \mathbb{R}$ by \begin{align*} U := \bigcup_{J \in \mathcal{U}} J. \end{align*} Since $\mathcal{U} \subseteq \mathcal{B}$ and $\mathcal{B}$ is countable, enumerate $\mathcal{U}$ as $(J_m)_{m \in M}$, where $M \subseteq \mathbb{N}$. For each $m \in M$, the set $J_m \setminus A$ is meagre in $J_m$. Thus there exists a meagre set $N_m \subseteq \mathbb{R}$ such that $J_m \setminus A \subseteq N_m$. Define \begin{align*} N_U := \bigcup_{m \in M} N_m. \end{align*} Because a countable union of meagre subsets of $\mathbb{R}$ is meagre, $N_U$ is meagre in $\mathbb{R}$. Also, \begin{align*} U \setminus A \subseteq N_U. \end{align*} Hence $U \setminus A$ is meagre. [/step]

[step:Show that $A$ is meagre outside the open set $U$]Let $F := \mathbb{R} \setminus U$. Let $\partial U$ denote the topological boundary of $U$ in the usual topology on $\mathbb{R}$, and let $\operatorname{int}(F)$ denote the topological interior of $F$. We prove that $A \cap F$ is meagre. First, the boundary $\partial U$ is nowhere dense in $\mathbb{R}$, since $U$ is open. Hence $A \cap \partial U$ is meagre. It remains to treat $A \cap \operatorname{int}(F)$. Let $K \in \mathcal{B}$ satisfy $K \subseteq \operatorname{int}(F)$. Since $K$ is not contained in $U$, no basic subinterval $J \in \mathcal{B}$ with $J \subseteq K$ can belong to $\mathcal{U}$. Apply the determined game dichotomy to $G_K(A)$. If player I had a winning strategy, the dichotomy would give $J \in \mathcal{B}$ with $\overline{J} \subset K$ and $J \setminus A$ meagre in $J$, so $J \in \mathcal{U}$ and therefore $J \subseteq U$, contradicting $K \subseteq F$. Therefore player I has no winning strategy in $G_K(A)$. By determinacy, player II has a winning strategy in $G_K(A)$. The dichotomy then gives that $A$ is meagre in $K$. Because $\operatorname{int}(F)$ is open, it is the union of the countable family \begin{align*} \mathcal{K} := \{K \in \mathcal{B} : K \subseteq \operatorname{int}(F)\}. \end{align*} For each $K \in \mathcal{K}$, the set $A \cap K$ is meagre in $K$, hence is contained in a meagre subset of $\mathbb{R}$. Since $\mathcal{K}$ is countable, $A \cap \operatorname{int}(F)$ is meagre in $\mathbb{R}$. Combining this with the meagreness of $A \cap \partial U$ gives that $A \cap F = A \setminus U$ is meagre.[/step]

[guided]The open set $U$ was built to include every basic interval on which $A$ is comeagre. We now show that this leaves only a meagre part of $A$ outside $U$. Set $F := \mathbb{R} \setminus U$. Since $U$ is open, every point of $F$ lies either in the boundary $\partial U$ or in the interior $\operatorname{int}(F)$. The boundary of an open subset of $\mathbb{R}$ is nowhere dense, so $A \cap \partial U$ is meagre. Thus the only non-immediate part is $A \cap \operatorname{int}(F)$. Take a basic interval $K \in \mathcal{B}$ with $K \subseteq \operatorname{int}(F)$. We apply the game dichotomy to $G_K(A)$. The game is determined by $\mathrm{AD}$. If player I had a winning strategy, the Banach-Mazur category dichotomy would produce a smaller basic interval $J \in \mathcal{B}$ with $\overline{J} \subset K$ such that $J \setminus A$ is meagre in $J$. But that condition is exactly the defining condition for $J$ to belong to $\mathcal{U}$. Hence $J \subseteq U$, contradicting $J \subseteq K \subseteq F$. Therefore player I cannot have a winning strategy. Since the game is determined, player II must have a winning strategy. Applying the second half of the Banach-Mazur category dichotomy gives that $A$ is meagre in $K$. Finally, $\operatorname{int}(F)$ is open, and every open subset of $\mathbb{R}$ is the union of countably many rational basic intervals contained in it. Hence \begin{align*} \operatorname{int}(F) = \bigcup \{K \in \mathcal{B} : K \subseteq \operatorname{int}(F)\}. \end{align*} On each such $K$, the set $A \cap K$ is meagre. A countable union of meagre sets is meagre, so $A \cap \operatorname{int}(F)$ is meagre. Together with the meagre boundary contribution, this proves that $A \setminus U = A \cap F$ is meagre.[/guided]

[step:Conclude that $A$ differs from an open set by a meagre set] We have constructed an open set $U \subseteq \mathbb{R}$ such that both $U \setminus A$ and $A \setminus U$ are meagre in $\mathbb{R}$. Therefore \begin{align*} A \triangle U = (A \setminus U) \cup (U \setminus A) \end{align*} is meagre as a countable union of two meagre sets. Hence $A$ has the Baire property. Since $A \subseteq \mathbb{R}$ was arbitrary, every set of reals has the Baire property under $\mathrm{AD}$. [/step]