Perfect-set game strategy dichotomy on Cantor space

Enumerate all finite binary words by length and then lexicographic order, including the empty word first. This gives I's natural-number codes. II's numbers zero and one are legal bits; at the first other number declare I the winner. At each legal round at least one output bit is appended, so the concatenation is infinite. A winning strategy on the coded game never prescribes a first illegal move on its consistent legal histories, since the opponent can always continue legally; restriction thus gives the stated game.

Fix an I winning strategy $\sigma$. For $z\in \mathcal{C}$ let $F(z)$ be the outcome against II's successive bits $z(n)$. It lies in $A$ by F2. If $z,w$ first differ at $n$, their game histories through I's nth block are identical, and their next bits differ at the same output position, so $F(z)\mathrm{\ne }F(w)$. If two inputs agree in their first $m$ bits, the first $m$ full rounds agree and append at least $m$ bits, hence their outputs agree in their first $m$ bits. Thus $F$ is continuous.

Fix instead a winning II strategy $\tau$ and $x\in A$. A barrier for $x$ is a finite legal full history $h$ consistent with $\tau$, ending before an I move, whose concatenation $s$ is a prefix of $x$, such that for every finite block $t$ with $s^{⌢}t\subset x$, the response $\tau (h^{⌢}t)$ differs from the next bit of $x$. If no such barrier existed, start with the empty history and at each round choose the least block code preserving agreement with $x$ after $\tau$ responds. The absence of a barrier makes this set nonempty at each resulting history. Recursion constructs a full $\tau$-play concatenating to $x$ (lengths grow by at least one), contrary to $x\in A$. Hence a barrier exists.

By F1 and the open-cover definition, the image of $\mathcal{C}$ under $F$ is compact: pull a cover back, take a finite subcover, and push its coverage forward. A compact set $K$ in a metric space is closed: for $x\mathrm{\notin }K$ the balls $B(y,d(x,y)\mathrm{/}3)$ cover $K$; finitely many suffice, and the minimum of the finitely many positive radii gives a ball about $x$ missing all those balls. Closed subsets of a compact space are compact, since adjoining their open complement to a cover gives a cover of the whole space. Consequently images under $F$ of closed subsets of $\mathcal{C}$ are compact and closed; injectivity now says its inverse on the image takes inverse images of closed sets to closed sets. Thus $F$ is a homeomorphism onto its image. That image has no isolated point because $\mathcal{C}$ has none by F1.

A fixed barrier history $h$ can serve at most one $x$. Its concatenation gives the first $\mathrm{\mid }s\mathrm{\mid }$ bits. Recursively, after reconstructing the additional block $t=x\upharpoonright [\mathrm{\mid }s\mathrm{\mid },\mathrm{\mid }s\mathrm{\mid }+j)$, recover the next bit as $1-\tau (h^{⌢}t)$. Each query uses the same fixed history $h$, not an evolving hypothetical history. The barrier property justifies every such recovered bit, including $j=0$ with empty $t$. Full finite histories have natural-number codes by iterated finite-word coding F1. Assign each $x\in A$ the least code of its barriers. Existence follows from step 1.3 and uniqueness per code from this reconstruction, so this assignment is an injection into $\mathbb{N}$. When $A=\mathrm{\varnothing }$ it is the empty injection. QED.