On Page 141, Axiom of Choice, Herrlich(2006)
Show that if in a game of the form $G(1, X_1, Y_1, A)$, the first player has no winning strategy, then the second player can always win, even though he may not have a winning strategy.
In this question, $G(1, X_1, Y_1, A)$ denotes a sequential game with two players. The first player starts with a choice from her action space $X_1$, and the second player player follows with a choice in $Y_1$. $A$ is a sub set of $X_1 \times Y_1$, if the outcome belongs to $A$, then the second player wins, otherwise, the first player wins. A winning strategy for the first player is a choice $x$ from $X_1$, such that $\{x\} \times Y_1 \bigcap A = \emptyset$ ,while a winning strategy for the second player is a function $f:X_1 \to Y_1$, such that $X_1 \times f(X_1) \subseteq A$.
I'm confused with the scenario that the second player always win without a winning strategy. Why not they are tautological?