**Proposition**: For each $y \in Y$, there exists an open set $N(y) \subset Y$ containing $y$ such that $X \times N(y)$ can be covered by a finite subfamily of $\mathcal{U}$ (the set $X \times N(y)$ is often called the tube about $X \times \{ y \}$) **Proof**: For each $x \in X$, we can find $W_x \in \mathcal{U}$ such that $(x, y) \in W_x$. By the definition of the [[Product topology|product topology]] there exist $U_x \in \mathcal{T}_X, V_x \in \mathcal{T}_Y$ such that $(x, y) \in U_x \times V_x \subset W_x .$ The collection $\left\{U_x: x \in X\right\}$ is an open cover of $X$, so it contains a finite subcover $\left\{U_{x_1}, \ldots U_{x_m}\right\}$. If we let $ N(y)=\bigcap_{i=1}^n V_{x_i} $ then $N(y) \subset Y$ is open, $y \in N(y)$ (so it is not empty), and $ X \times N(y) \subset \bigcup_{i=1}^n\left(U_{x_i} \times V_{x_i}\right) \subset \bigcup_{i=1}^n W_{x_i} $ This proves the claim.