Proof Notice first that B is not a Borel set: for if it were, either B or $X \setminus B$ would contain a non-empty perfect set. And since B is not Borel, the function $1_{B}$ is not a limsup function, and hence Player II has no winning strategy in $\Gamma (1_{B})$ .

Suppose that Player I does.

Let $E = \{v \in 2^{\mathbb {N}}:v_{n} = 1\text { for infinitely many }n\in \mathbb {N}\}$ , where we write $2 = \{0,1\}$ . Notice that E is not a $\mathbf {\Sigma }_{2}^{0}$ subset of $2^{\mathbb {N}}$ . For it were, we would be able to express $2^{\mathbb {N}}$ as a countable union of meagre sets, contradicting the Baire category theorem.

Now, consider the continuous function $g : 2^{\mathbb {N}} \to X$ induced by Player I’s winning strategy. Then, $g(E) \subseteq X \setminus B$ and $g(2^{\mathbb {N}} \setminus E) \subseteq B$ . Since $g(E)$ is an analytic subset of X, it is either countable, or it contains a perfect subset. But $X \setminus B$ contains no perfect subset. Thus $g(E)$ is countable, hence a $\mathbf {\Sigma }_{2}^{0}$ subset of X. But then, $E = g^{-1}(g(E))$ is a $\mathbf {\Sigma }_{2}^{0}$ subset of $2^{\mathbb {N}}$ , yielding a contradiction.

Proof Let $Y = C \cap \{f \ge r\}$ , and let $\{S_{0},S_{1},\dots \}$ be a cover of Y by closed nowhere dense subsets of C. We presently construct a winning strategy for Player I.

Fix some sequence $v_{0},v_{1},\dots $ of Player II’s moves.

Let $y(0)$ be any point of the set $Y \setminus S_{0}$ . Notice that the set $C \setminus S_{0}$ is not empty because $S_{0}$ is nowhere dense in C, and $Y \setminus S_{0}$ is not empty since Y is dense in C. Set $m_{0} = 0$ .

Player I starts with a move $x_{0} = y(0)_{0}$ . Take an $n \in \mathbb {N}$ and suppose that Player I’s moves $x_{0}, \dots , x_{n}$ have been defined, along with a point $y(n) \in Y$ and a number $m_{n} \in \mathbb {N}$ , such that

(3.1) $$ \begin{align} (x_{0},\dots,x_{n}) = (y(n)_0,\dots,y(n)_n). \end{align} $$

To define the next move of Player I, $x_{n+1}$ , we distinguish two cases:

Case 1: $v_{n}> r - 2^{-m_{n}}$ and $O(x_{0},\dots ,x_{n}) \cap S_{m_{n}} = \emptyset $ . Let $y(n+1)$ be any point of the set $(O(x_{0},\dots ,x_{n}) \cap Y) \setminus S_{m_{n}+1}$ . Notice that the set $(O(x_{0},\dots ,x_{n}) \cap C) \setminus S_{m_{n}+1}$ is not empty because $S_{m_{n}+1}$ is nowhere dense in C, and $(O(x_{0},\dots ,x_{n}) \cap Y) \setminus S_{m_{n}+1}$ is not empty since Y is dense in C. Let $m_{n+1} = m_{n} + 1$ , and define Player I’s move as $x_{n+1} = y(n+1)_{n+1}$ .

Case 2: otherwise. In this case we let $y(n+1) = y(n)$ , $m_{n+1} = m_{n}$ , and define Player I’s move as $x_{n+1} = y(n+1)_{n+1}$ .

Notice that in either case (3.1) holds for $n+1$ . This completes the definition of Player I’s strategy.

The intuition behind this definition could be explained as follows: Player I starts by zooming in on the point $y(0)$ chosen to be in Y but not in $S_{0}$ . Player I awaits a stage n where Player II would make a move $v_{n}> r - 1$ , and where the set $S_{0}$ would be “excluded.” As soon as such a stage is reached, Player I switches to an element $y(1)$ , chosen to be in Y but not in $S_{1}$ . He then zooms in on $y(1)$ , awaiting a stage where Player I would make a move $v_{n}> r - 1/2$ , and where $S_{1}$ would be “excluded.” As soon as such a stage occurs, Player I switches to an element $y(2)$ chosen to be in Y but not in $S_{2}$ . And so on.

We argue that Player I’s strategy is winning.

Suppose first that $\limsup v_{n} \leq r - 2^{-m}$ for some $m \in \mathbb {N}$ . Then, Case 1 occurs at most finitely many times. Let N be the last stage when Case 1 occurs (or $N = 0$ if Case 1 never occurs). Then, the point x produced by Player I equals to $y(N)$ . We thus have $\limsup v_{n} \leq r - 2^{-m} < r \leq f(y(N)) = f(x)$ .

Suppose now that $\limsup v_{n} \geq r$ . We argue that Case 1 occurs infinitely many times. Suppose to the contrary and let N be the last stage when Case 1 occurs (or $0$ if Case 1 never occurs). Then, $m_{N} = m_{N+1} = \cdots $ and $y(N) = y(N+1) = \cdots = x$ . There are infinitely many $n> N$ with $v_{n}> r - 2^{-m_{N}}$ , and for each such n the neighborhood $O(x_{0},\dots , x_{n})$ of x has a point in common with $S_{m_{N}}$ . This implies that $x \in S_{m_{N}}$ . This, however, contradicts the choice of $y(N)$ . This establishes that Case 1 occurs infinitely many times.

Let x be the point constructed by Player I. We argue that $x \in C \setminus Y$ . In view of (3.1), x is a limit of the sequence $y(0),y(1),\dots $ . Since each $y(n)$ is an element of the closed set C, so is x. To see that x is not an element of Y, suppose to the contrary. Then, $x \in S_{m}$ for some $m \in \mathbb {N}$ . Since Case 1 occurs infinitely often, the sequence $m_{0},m_{1},\dots $ runs through all natural numbers, so we can choose $n \in \mathbb {N}$ to be the largest number such that $m_{n} = m$ . This choice implies that Case 1 occurs at stage n, and hence $O(x_{0},\dots ,x_{n})$ is disjoint from $S_{m_{n}}$ , leading to a contradiction.

It follows that x is not an element of $\{f \ge r\}$ . Thus $\limsup v_{n} \geq r> f(x)$ , which completes the proof.

Proof It is straightforward to check that if Player I has a winning strategy in $\Gamma (f)$ , then the restriction of this strategy is winning for Player I in $\Gamma _R(f)$ .

Conversely, fix a winning strategy $\sigma _R$ of Player I for the game $\Gamma _R(f)$ . For each $n \in \mathbb {N}$ define $F_n: \mathbb {R} \to R$ so that

(3.2) $$ \begin{align} \forall y \in \mathbb{R}, \ n \in \mathbb{N}{:} \ |F_n(y) - y | < \text{d}(y,R) + \tfrac{1}{n} \end{align} $$

holds. Now define Player I’s strategy $\sigma $ in $\Gamma (f)$ as follows: let $\sigma (\emptyset ) = \sigma _R(\emptyset )$ , and let

(3.3) $$ \begin{align} \sigma(x_0,v_0,x_1,v_1, \dots, x_n,v_n) = \sigma_R(x_0, F_0(v_0), x_1, F_1(v_1), \dots, x_n, F_n(v_n)) \end{align} $$

whenever $n \in \mathbb {N}$ and $(x_0, \dots , x_n) \in T\!$ .

It remains to check that $\sigma $ is a winning strategy for Player I in $\Gamma (f)$ . Fix a run $x_0,v_0,x_1,v_1, \dots $ of the game $\Gamma (f)$ consistent with $\sigma $ , i.e., such that for each $n \in \mathbb {N}$ , $\sigma (x_0,v_0, \dots , x_n,v_n) = x_{n+1}$ . Then, (3.3) implies that for each n

(3.4) $$ \begin{align} \sigma_R(x_0, F_0(v_0), x_1, F_1(v_1), \dots, x_n, F_n(v_n)) = x_{n+1}, \end{align} $$

and as $\sigma _R$ is a winning strategy for Player I in $\Gamma _{R}(f)$ , we obtain that

$$ \begin{align*} f(x_0,x_1,x_2, \dots) \neq \limsup_{n \to \infty} F_n(v_n). \end{align*} $$

We have to check that $f(x_0,x_1,x_2, \dots ) \neq \limsup _{n \to \infty } v_n$ . First, if $\limsup _{n \to \infty } v_n \notin R \supseteq \text {ran}(f)$ , we are done. Otherwise $\limsup _{n \to \infty } v_n = r \in R$ , therefore for each $\varepsilon> 0$ , for all but finitely many k we have $v_k < r + \varepsilon $ , thus (3.2) implies $F_k(v_k) < r + 2\varepsilon + \frac {1}{k}$ for these cofinitely many k’s, therefore $\limsup _{k \to \infty } F_k(v_k) \leq r$ . Since $r - \varepsilon < v_k$ holds for infinitely many k, this argument also shows that $r - 2\varepsilon - \tfrac {1}{k} < F_k(v_k)$ for infinitely many k too, thus

$$ \begin{align*}\limsup_{n \to \infty} v_n = r = \limsup_{n \to \infty} F_n(v_n) \neq f(x_0,x_1,x_2, \dots),\end{align*} $$

as desired.

Proof Define R to be the closure of the range of f. Then, it is easy to verify that R contains no infinite strictly increasing sequence. Hence, the usual order $>$ of the reals is a well ordering of R. Let $\rho $ be the order type of $(R,>)$ , and let $\alpha \mapsto r_{\alpha }$ be the bijective map from $\rho $ to R such that $r_{\alpha }> r_{\beta }$ whenever $\alpha < \beta $ . Notice that $\rho $ is a countable ordinal.

Assume that Player I has a winning strategy in $\Gamma (f)$ . Then, by Lemma 3.5 Player I also has a winning strategy in $\Gamma _R (f)$ . Let $\sigma _R$ be such a strategy. Let $g : R^{\mathbb {N}} \to X$ be the continuous function induced by $\sigma _R$ . Here R is given its discrete topology; since R is countable, $R^{\mathbb {N}}$ is a Polish space. For each $r \in R$ let $L_{r} = \{v \in R^{\mathbb {N}}: \limsup _{t \to \infty }v_{t} = r\}$ , and let $A_{r} = g(L_{r})$ . The set $A_{r}$ is analytic. Moreover,

(3.5) $$ \begin{align} \{f = r\} \cap A_{r} = \emptyset. \end{align} $$

Suppose that the function f fails to satisfy the conclusion of the theorem, that is, there is no number $r\in \mathbb {R}$ and Cantor set $C\subseteq X$ such that $C\cap \{f\ge r\}$ is countable and dense in C. We obtain a contradiction by showing that there exists a limsup function $e : X \to R$ such that Player I has a winning strategy in the game $\Gamma (e)$ . More precisely, we show that $\sigma _R$ is a winning strategy for Player I in $\Gamma _R(e)$ , which suffices by Lemma 3.5.

Note that a function $e : X \to R$ is a limsup function if and only if $\{e \ge r\}$ is a $\mathbf {\Pi }_{2}^{0}$ set for each $r \in R$ , using that R is closed.

We define recursively a sequence $(G_{\alpha }:\alpha < \rho )$ of $\mathbf {\Pi }_{2}^{0}$ subsets of X such that $\{f \ge r_\alpha \} \subseteq G_{\alpha }$ . Let $\beta < \rho $ be an ordinal such that the sets $(G_{\alpha }:\alpha < \beta )$ have been defined. In particular, notice that

(3.6) $$ \begin{align} \{f> r_{\beta}\} \subseteq \bigcup_{\alpha < \beta}\, \bigcap_{\gamma:\, \alpha \leq \gamma <\beta} G_{\gamma}. \end{align} $$

Since f fails to satisfy the condition of the theorem, there exists no Cantor set $C \subseteq X$ such that $C \cap \{f \ge r_\beta \}$ is countable and dense in C. This implies (using [Reference Kechris6, Theorem 21.22]) that $\{f \ge r_\beta \}$ can be separated from any disjoint analytic subset of X by a $\mathbf {\Pi }_{2}^{0}$ set. Consider the set

(3.7) $$ \begin{align} A_{r_{\beta}} \Big\backslash \bigcup_{\alpha < \beta}\, \bigcap_{\gamma:\, \alpha \leq \gamma <\beta} G_{\gamma}. \end{align} $$

It is analytic, since $\beta $ is a countable ordinal. Moreover, it is disjoint from $\{f \ge r_\beta \}$ as can be seen from (3.5) and (3.6). Hence there exists a $\mathbf {\Pi }_{2}^{0}$ subset $G_{\beta }$ of X containing $\{f \ge r_\beta \}$ and disjoint from (3.7). Thus $\{f \ge r_\beta \} \subseteq G_{\beta }$ and

(3.8) $$ \begin{align} G_{\beta} \cap A_{r_{\beta}} \subseteq \bigcup_{\alpha < \beta}\, \bigcap_{\gamma:\, \alpha \leq \gamma <\beta} G_{\gamma}. \end{align} $$

This concludes the recursive definition of the sequence $(G_\alpha : \alpha <\rho )$ .

Now, for an arbitrary $\beta < \rho $ define the set

$$ \begin{align*}E_{\beta} = \bigcap_{\gamma:\, \beta \leq \gamma < \rho} G_{\gamma}.\end{align*} $$

This is a $\mathbf {\Pi }_{2}^{0}$ set, since $\rho $ is a countable ordinal. Moreover, $\{f \ge r_\beta \} \subseteq E_{\beta }$ for each $\beta < \rho $ , hence $X = \bigcup _{\beta <\rho }E_{\beta }$ . In view of (3.8), we have

(3.9) $$ \begin{align} E_{\beta} \cap A_{r_{\beta}} \subseteq \bigcup_{\alpha < \beta} E_{\alpha}. \end{align} $$

Define $e \colon X \to R$ by letting $e(x) = r_{\beta }$ , where $\beta < \rho $ is the least ordinal such that $x \in E_{\beta }$ . Then, $\{e \ge r_\beta \} = E_{\beta }$ for each $\beta <\rho $ , so e is a limsup function. Moreover, $\{e = r_{\beta }\}$ equals the set $E_{\beta } \setminus \bigcup _{\alpha < \beta }E_{\alpha }$ , which is disjoint from $A_{r_{\beta }}$ by (3.9). This shows that Player I’s strategy $\sigma _R$ remains winning in the game $\Gamma _{R}(e)$ , yielding the desired contradiction.

Proof If for each $r \in \mathbb {R}$ , $\{f \ge r \}$ is a $\mathbf {\Pi }_{2}^{0}$ set, then f is a limsup function, hence Player II has a winning strategy in $\Gamma (f)$ , a contradiction. Hence $\{f \ge r \}$ is not a $\mathbf {\Pi }_{2}^{0}$ set for some $r \in \mathbb {R}$ . Then, the Hurewicz theorem (see, e.g., [Reference Kechris6, Theorem 21.18]) implies that there is a Cantor set C such that the set $C \cap \{f \ge r\}$ is countable and dense in C.

Proof If for each $r \in \mathbb {R}$ , $\{f \ge r \}$ is a $\mathbf {\Pi }_{2}^{0}$ set, then f is a limsup function, hence Player II has a winning strategy in $\Gamma (f)$ . Otherwise, $\{f \ge r \}$ is not a $\mathbf {\Pi }_{2}^{0}$ set for some $r \in \mathbb {R}$ , and as above, the Hurewicz theorem implies that there is a Cantor set C such that the set $C \cap \{f \ge r\}$ is countable and dense in C, therefore Player I has a winning strategy by Theorem 3.3.

Proof First note that every Cantor set can be written as a disjoint union of uncountably many (in fact, continuum many) Cantor sets, since it is well-known that a Cantor set is homeomorphic to $2^I$ for every countably infinite set I, in particular to $2^{\mathbb {N} \times \mathbb {N}}$ , which is homeomorphic to $2^{\mathbb {N}} \times 2^{\mathbb {N}} = \bigcup _{c \in 2^{\mathbb {N}}} \left (\{c\} \times 2^{\mathbb {N}} \right )$ .

This implies that if H is an arbitrary set, then in order to show that $C \cap H$ is uncountable for each Cantor set $C \subseteq \mathbb {N}^{\mathbb {N}}$ , it suffices to show that $C \cap H \neq \emptyset $ for each Cantor set $C \subseteq \mathbb {N}^{\mathbb {N}}$ .

Let $X = \mathbb {N}^{\mathbb {N}}$ and let $\varphi : X \to \mathbb {N} \cup \{+\infty \}$ be given by $\varphi (x) = \limsup _{n \to \infty }x_{n}$ . We first argue that there exists a function $f : X \to \mathbb {N}$ such that (a) $f(x) \neq \varphi (x)$ for each $x \in X$ , and (b) $C \cap \{f \ge r\} \neq \emptyset $ for each $r \in \mathbb {R}$ and each Cantor set $C \subseteq X$ . We will then show that condition (a) implies that Player I has a winning strategy in $\Gamma (f)$ , while we already argued that (b) implies that $C \cap \{f \ge r\}$ is uncountable for each $r \in \mathbb {R}$ and each Cantor set $C \subseteq \mathbb {N}^{\mathbb {N}}$ .

Let $(r_{\alpha }:\alpha <\mathfrak {c})$ , $(z_{\alpha }:\alpha <\mathfrak {c})$ , and $(C_{\alpha }:\alpha <\mathfrak {c})$ be enumerations of the real numbers, of the points of X, and of the Cantor subsets of X, respectively. We define the pairs $(z_{\alpha }, f(z_{\alpha })) \in X \times \mathbb {N}$ recursively as follows. Take an ordinal $\alpha < \mathfrak {c}$ and suppose that $(z_{\beta },f(z_{\beta }))$ has been defined for every $\beta < \alpha $ . Let $z_{\alpha }$ be any point of $C_{\alpha } \setminus \{z_{\beta }:\beta <\alpha \}$ . Define $f(z_{\alpha })$ to be the smallest natural number such that $f(z_{\alpha }) \geq r_{\alpha }$ and $f(z_{\alpha }) \neq \varphi (z_{\alpha })$ . To complete the definition of f, for each point $x \in X \setminus \{z_{\beta }:\beta <\mathfrak {c}\}$ let $f(x)$ be the smallest natural number such that $f(x) \neq \varphi (x)$ .

Now we show that Player I has a winning strategy in $\Gamma (f)$ . Using Lemma 3.5, it is enough to show that Player I has a winning strategy in $\Gamma _{\mathbb {N}}(f)$ . Let Player I start by playing $x_{0} = 0$ . To a move $v_{n} \in \mathbb {N}$ of Player II in round n, Player I responds with $x_{n+1} = v_n$ . Then, for a run $x_{0},v_{0},x_{1},v_{1},\dots $ of the game, it holds that $\varphi (x) = \limsup _{n \to \infty }x_{n} = \limsup _{n \to \infty }v_n$ . Since $f(x) \neq \varphi (x)$ holds, we have $f(x) \neq \limsup _{n \to \infty } v_n $ , hence the run is won by Player I.

Proof Let $i : \mathbb {N} \to R$ be a strictly increasing map. Let $f_0 : \mathbb {N}^{\mathbb {N}} \to \mathbb {N}$ be a function as in Theorem 3.9, that is, such that Player I has a winning strategy in $\Gamma (f_0)$ , and $C \cap \{f_0 \ge r\}$ is uncountable for each $r \in \mathbb {R}$ and each Cantor set $C \subseteq \mathbb {N}^{\mathbb {N}}$ . We claim that the function defined as $f = i \circ f_0$ works. Clearly, $f : \mathbb {N}^{\mathbb {N}} \to R$ , and it is also clear that $C \cap \{f \ge r\}$ is either uncountable or empty for each $r \in \mathbb {R}$ and each Cantor set $C \subseteq \mathbb {N}^{\mathbb {N}}$ , hence we only have to show that Player I has a winning strategy in $\Gamma (f)$ . By Lemma 3.5 it suffices to check that Player I has a winning strategy in $\Gamma _{i(\mathbb {N})}(f)$ . Let $\sigma _0$ be a winning strategy for Player I in $\Gamma (f_0)$ , and define for each $n \in \mathbb {N}$

(3.10) $$ \begin{align} \sigma_{i (\mathbb{N} )} (x_0,v_0, \dots, x_n,v_n) = \sigma_0(x_0, i^{-1}(v_0), \dots, x_n, i^{-1}(v_n)). \end{align} $$

Since i is order-preserving, it is easy to check that $\sigma _{i (\mathbb {N} )}$ is a winning strategy for Player I in $\Gamma _{i(\mathbb {N})}(f)$ .

Proof Let $X = \mathbb {N}^{\mathbb {N}}$ . Let $q(0), q(1), \dots $ be an enumeration of the rational numbers, and let $\varphi : X \to \mathbb {R}\cup \{+\infty \}$ be given by $\varphi (x) = \limsup _{n \to \infty } q(x_n)$ . In order to construct $f : X \to \mathbb {R}$ with co-analytic graph, we use a result of Vidnyánszky [Reference Vidnyánszky13, Theorem 1.3]. Let $B_1 = \{ (C, t) \in \mathcal {K}(X) \times \mathbb {R} : C$ is a Cantor set $\}$ , where $\mathcal {K}(X)$ is the family of non-empty compact sets in X equipped with the Hausdorff metric. Let $B_2 = \mathbb {R}$ and $B = B_1 \sqcup B_2$ be the disjoint union of $B_1$ and $B_2$ making B a subset of the Polish space $(\mathcal {K}(X)\times \mathbb {R}) \sqcup \mathbb {R}$ . Let $i : X \to \mathbb {R}$ be a Borel bijection, $M = \mathbb {R}^2$ , and let

$$ \begin{align*}F_1 = &\Big\{\big(A, (C, r), (y, t)\big) \in M^{\le \omega} \times B_1 \\&\quad\times M : y \in i(C) \setminus \textrm{pr}_1(\textrm{ran}(A)), t \ge r, t \neq \varphi(i^{-1}(y))\Big\},\end{align*} $$

where $\textrm {pr}_1(\textrm {ran}(A))$ is the projection of the range of the sequence A onto the first coordinate. Let

$$ \begin{align*} F_2 = \Big\{\!\big(A, y', (y, t)\big) \in M^{\le \omega} \times B_2 \times M : \; & t \neq \varphi\big(i^{-1}(y)\big), y' \not\in \textrm{pr}_1\big(\textrm{ran}(A)\big) \kern1.5pt{\Rightarrow}\kern1.5pt y' \kern1.5pt{=}\kern1.5pt y, \\ & y' \in \textrm{pr}_1\big(\textrm{ran}(A)\big) \Rightarrow y \not\in \textrm{pr}_1\big(\textrm{ran}(A) \big)\Big\}, \end{align*} $$

and let $F = F_1 \sqcup F_2 \subseteq M^{\le \omega } \times B \times M$ .

We now check that the conditions of Vidnyánszky’s theorem are satisfied. First, a non-empty compact set $C \subseteq \mathbb {N}^{\mathbb {N}}$ is a Cantor set if and only if it is perfect. Using [Reference Kechris6, Exercise 4.31] one can easily see that $B_1$ is a Borel subset of $\mathcal {K}(X) \times \mathbb {R}$ . Therefore B is a Borel subset of $(\mathcal {K}(X) \times \mathbb {R}) \sqcup \mathbb {R}$ . The set $F_1$ is clearly co-analytic, and since $A \in M^{\le \omega }$ is a countable sequence, conditions of the form $y' \in \textrm {pr}_1(\textrm {ran}(A))$ are Borel. Therefore $F_2$ is even Borel, making $F = F_1 \sqcup F_2$ co-analytic. For each $(A, b) \in M^{\le \omega } \times B$ , no matter whether $b \in \mathcal {K}(X) \times \mathbb {R}$ or $b \in \mathbb {R}$ , the section

$$ \begin{align*}F_{(A, b)} = \{(y, t) \in M : \big(A, b, (y, t)\big) \in F\}\end{align*} $$

contains $\{x_1\} \times \{t : t \ge x_2\}$ for some $(x_1, x_2) \in \mathbb {R}^2$ , hence it is cofinal in the Turing degrees (for this notion, see Definition 1.1 of [Reference Vidnyánszky13]). Therefore the conditions of the theorem are satisfied.

The conclusion of the theorem assures that there is a co-analytic set $G \subseteq M = \mathbb {R}^2$ and enumerations $B = \{b_\alpha : \alpha < \omega _1\}$ , $G = \{g_\alpha : \alpha < \omega _1\}$ and for every $\alpha < \omega _1$ a sequence $A_\alpha \in M^{\le \omega }$ that is an enumeration of $\{g_\beta : \beta < \alpha \}$ such that $g_\alpha \in F_{(A_\alpha , b_\alpha )}$ for every $\alpha < \omega _1$ . We note here that the assumption $V = L$ implies the continuum hypothesis.

First we check that G is the graph of a function with domain $\mathbb {R}$ . Notice that for $\beta < \alpha $ , if $g_\alpha = (y_1, t_1)$ and $g_\beta = (y_2, t_2)$ , then $y_1 \neq y_2$ . Indeed, $g_\alpha \in F_{(A_\alpha , b_\alpha )}$ implies that $y_1 \not \in \textrm {pr}_1(\textrm {ran}(A_\alpha ))$ , and since $y_2 \in \textrm {pr}_1(\textrm {ran}(A_\alpha ))$ , $y_1 \neq y_2$ easily follows. To see that for each $y \in \mathbb {R}$ , $(y, t) \in G$ for some $t \in \mathbb {R}$ , let $\alpha < \omega _1$ be chosen with $b_\alpha = y \in B_2$ . Then, either $y \in \textrm {pr}_1(\textrm {ran}(A_\alpha ))$ and we are done, or $g_\alpha $ is chosen to be $(y, t)$ for some $t \in \mathbb {R}$ . Therefore G is indeed a graph of a function with domain $\mathbb {R}$ .

Now we define the function $f : X \to \mathbb {R}$ in the following way: for each $(y, t) \in G$ , let $f(i^{-1}(y)) = t$ . Clearly, the graph of f is $(i, \textrm {id})^{-1}(G)$ , hence it is co-analytic.

We now show that the defined function f has properties (a) $f(x) \neq \varphi (x)$ for each $x \in X$ , and (b) $C \cap \{f \ge r\} \neq \emptyset $ for each $r \in \mathbb {R}$ and each Cantor set $C \subseteq X$ . Then, we will show that (a) implies that Player I has a winning strategy in $\Gamma (f)$ . The proof that (b) implies that $C \cap \{f \ge r\}$ is uncountable for each $r \in \mathbb {R}$ and each Cantor set C is exactly the same as in the proof of Theorem 3.9.

To show (a), let $(x, t) \in X \times \mathbb {R}$ be a pair with $(i(x), t) = g_\alpha \in G$ . Then, $g_\alpha \in F_{(A_\alpha , b_\alpha )}$ implies $t \neq \varphi (x)$ , hence $f(x) = t \neq \varphi (x)$ . To show (b), let $C \subseteq X$ be a Cantor set and let $r \in \mathbb {R}$ . Let $\alpha < \omega _1$ be the ordinal with $b_\alpha = (C, r)$ . Then, for $g_\alpha = (y, t)$ , using again that $g_\alpha \in F_{(A_\alpha , b_\alpha )}$ , $y \in i(C)$ and $t \ge r$ , hence $i^{-1}(y) \in C$ and $f(i^{-1}(y)) = t \ge r$ .

It remains to show that Player I has a winning strategy in $\Gamma (f)$ . Let Player I start by playing $x_{0} = 0$ . To a move $v_{n}$ of Player II, Player I responds with an $x_{n+1} \in \mathbb {N}$ chosen to be the smallest natural number satisfying $|v_{n} - q(x_{n+1})| \leq 2^{-n}$ . Then, for a run $x_{0},v_{0},x_{1},v_{1},\dots $ of the game, it holds that $\varphi (x) = \limsup _{n \to \infty }v_{n}$ . Since $f(x) \neq \varphi (x)$ , the run is won by Player I.

Question 3.12. For which $f : X \to \mathbb {R}$ does Player I have a winning strategy in $\Gamma (f)$ ?

Proof Assume that $\mathrm {osc}_f(C,x) \geq 5\epsilon> 0$ for each $x \in C$ . We will first describe a strategy of Player I and then we will show that it is a winning strategy. To define the moves of Player I in a particular run, we will use recursion to define natural numbers $n_0<n_1<n_2<\cdots $ and sequences $s_0, s_1, s_2, \ldots \in T$ (these may depend on the moves of Player II).

Let $n_{0} = 0$ , and let $s_{0}$ be the empty sequence. Suppose that, for some even number $k \in \mathbb {N}$ , Player I’s moves prior to the stage $n_{k}$ have been defined.

Let $s_{k} \in T$ denote the sequence of Player I’s moves prior to the stage $n_{k}$ . Define $\alpha _{k} = \sup \{f(x): x \in O(s_{k}) \cap C\}$ , and choose a point $x(k) \in O(s_{k}) \cap C$ so that $\alpha _{k} - \epsilon < f(x(k))$ . Starting with the stage $n_{k}$ , Player I produces his moves using the point $x(k)$ , that is, he plays $x_{n} = x(k)_n$ at a stage $n \geq n_{k}$ . He continues doing so until the first stage, say $n_{k+1}> n_{k}$ , that Player II makes a move $(v_{n_{k+1}}, w_{n_{k+1}})$ such that $|v_{n_{k+1}} - f(x(k))| < \epsilon $ . If no such stage occurs, then Player I goes on using the point $x(k)$ to make his moves until the end of the game.

Let $s_{k+1} \in T$ denote the sequence of moves produced by Player I prior to the stage $n_{k+1}$ . Define $\beta _{k+1} = \inf \{f(x): x \in O(s_{k+1}) \cap C\}$ , and choose a point $x(k+1) \in O(s_{k+1}) \cap C$ so that $f(x(k+1)) < \beta _{k+1} + \epsilon $ . Starting with the stage $n_{k+1}$ , Player I produces the moves using $x(k+1)$ , that is he plays $x_{n} = x(k+1)_n$ at a stage $n \geq n_{k+1}$ . He continues doing so until the first stage, say $n_{k+2}> n_{k+1}$ , that Player II makes a move $(v_{n_{k+2}},w_{n_{k+2}})$ such that $|w_{n_{k+2}} - f(x(k+1))| < \epsilon $ . If no such stage occurs, then Player I goes on using the point $x(k+1)$ until the end of the game.

We show that the strategy thus defined is winning.

Suppose first that only finitely many stages $n_{0},n_{1},\dots $ occur, the last one being $n_{k}$ . For concreteness, suppose that k is even. In this case Player I uses the point $x(k)$ to generate his moves until the end of the game. Moreover, there is no $n> n_{k}$ such that $|v_{n} - f(x(k))| < \epsilon $ . This implies that $\limsup v_{n} \neq f(x(k))$ , and hence the run is won by Player I. Likewise, if the last one of the sequence $n_{0},n_{1},\dots $ is the stage $n_{k+1}$ where k is even, then Player I generates the point $x(k+1)$ , and there is no $n> n_{k+1}$ such that $|w_{n} - f(x(k+1))| < \epsilon $ . Therefore $\liminf w_{n} \neq f(x(k+1))$ , and hence the run is won by Player I.

Suppose that infinitely many stages $n_{0},n_{1},\dots $ occur. From the above definitions we get for each even $k \in \mathbb {N}$

$$ \begin{align*} v_{n_{k+1}} &> f(x(k)) - \epsilon\\ &> \alpha_{k} - 2\epsilon\\ &=(\alpha_{k} - \beta_{k+1}) + \beta_{k+1} - 2\epsilon\\ &\geq (\alpha_{k} - \beta_{k+1}) + f(x(k+1)) - 3\epsilon\\ &\geq (\alpha_{k} - \beta_{k+1}) + w_{n_{k+2}} - 4\epsilon. \end{align*} $$

Let $\alpha _{k+1} = \sup \{f(x):x \in C \cap O(s_{k+1})\}$ . Since the sequence $s_{k+1}$ extends $s_{k}$ , we have $\alpha _{k} \geq \alpha _{k+1}$ . By the assumption, the oscillation of $f|_{C}$ at the point $x(k+1) \in C$ is at least $5\epsilon $ , hence $\alpha _{k+1} - \beta _{k+1} \geq 5\epsilon $ . Combining these facts we obtain that for each even $k \in \mathbb {N}$ it holds that $v_{n_{k+1}} \geq w_{n_{k+2}} + \epsilon $ . This, however, means that $\limsup v_{n}> \liminf w_{n}$ , implying a win for Player I.

Proof If f is a function in Baire class 1, then Player II has a winning strategy by Lemma 4.1. Suppose that f is not a function in Baire class 1. Then (see [Reference Kuratowski9, Theorem 2 and Remark 1, p. 395]) there exists a non-empty closed set $K \subseteq X$ such that the set of discontinuity points of $f|_{K}$ contains an open subset of K. Using the Baire category theorem and the arguments as in [Reference Kiss8, p. 9] one can show that there is a non-empty closed set $C \subseteq K$ such that the oscillation of $f|_{C}$ is bounded away from zero. The preceding lemma then implies that Player I has a winning strategy.