Romania — NMO Selection Tests for the Balkan and International Mathematical Olympiads

Let $S(X, Y) = \{x + y - \lfloor x + y \rfloor : x \in X \text{ and } y \in Y\}$ and proceed by induction on $|Y|$.

If $|Y| = 1$, the statement is clear.

Assume henceforth $|Y| > 1$ and let $y_0 \in Y$, $y_0 \neq 0$. The conditions $y_0 \neq 0$ and $x + y_0 = 1$ for no $x \in X$ imply that $0 \notin S(X, y_0)$. Since $0 \in X$, and $X$ and $S(X, y_0)$ have the same cardinality, there are elements that lie in $S(X, y_0)$ but not in $X$; that is, $x_0 + y_0 - \lfloor x_0 + y_0 \rfloor$ for some $x_0 \in X$.

Let
$$Y_0 = \{y : y \in Y \text{ and } x_0 + y - \lfloor x_0 + y \rfloor \notin X\}$$
and
$$X_0 = \{x_0 + y - \lfloor x_0 + y \rfloor : y \in Y_0\}.$$
Clearly, $0 \notin Y_0$, $X_0$ and $Y_0$ are both non-empty and have the same cardinality, so $0 < |X_0| = |Y_0| < |Y|$.

Let further $X' = X \cup X_0$ (notice that this is a disjoint union) and $Y' = Y \setminus Y_0$, so $X$ is a proper non-empty subset of $X'$, $Y'$ is a proper non-empty subset of $Y$, and
$$|X'| + |Y'| = |X| + |X_0| + |Y| - |Y_0| = |X| + |Y|. \quad (*)$$
Obviously, $X'$ and $Y'$ both contain $0$ and we now show that $S(X', Y') \subseteq S(X, Y)$ and $x' + y' = 1$ for no $x' \in X'$ and no $y' \in Y'$.

Clearly, we need consider only the case $x' \in X_0$; that is, $x' = x_0 + y - \lfloor x_0 + y \rfloor$ for some $y \in Y_0$. Thus, if $y' \in Y'$, then
$$\begin{align*} x' + y' - \lfloor x' + y' \rfloor &= x_0 + y - \lfloor x_0 + y \rfloor + y' - \lfloor x_0 + y - \lfloor x_0 + y \rfloor + y' \rfloor \\ &= x_0 + y + y' - \lfloor x_0 + y + y' \rfloor \\ &= (x_0 + y' - \lfloor x_0 + y' \rfloor) + y - \lfloor (x_0 + y' - \lfloor x_0 + y' \rfloor) + y \rfloor . \end{align*}$$
Notice that $x_0 + y' - \lfloor x_0 + y' \rfloor \in X$, by the definition of $Y'$, to infer that $x' + y' - \lfloor x' + y' \rfloor \in S(X, Y_0) \subseteq S(X, Y)$, and thereby $S(X', Y') \subseteq S(X, Y)$.

Finally, write
$$\begin{align*} x' + y' &= x_0 + y - \lfloor x_0 + y \rfloor + y' \\ &= (x_0 + y' - \lfloor x_0 + y' \rfloor) + y - \lfloor x_0 + y \rfloor + \lfloor x_0 + y' \rfloor, \end{align*}$$
recall that $x_0 + y' - \lfloor x_0 + y' \rfloor \in X$ and consider the possible values of $\lfloor x_0 + y \rfloor$ and $\lfloor x_0 + y' \rfloor$ to conclude that $x' + y' \neq 1$.

Consequently,
$$\begin{align*} |S(X, Y)| &\ge |S(X', Y')| && \text{for } S(X', Y') \subseteq S(X, Y) \\ &\ge |X'| + |Y'| - 1 && \text{by the induction hypothesis} \\ &= |X| + |Y| - 1. && \text{by } (*) \end{align*}$$