Let $(X, \mathcal{T}_{X}), (Y, \mathcal{T})_{Y}$ be [[Topology|topological spaces]]. Recall that the [[Product Topology|product topology]] on $X \times Y$ is the topology with base $ \left\{U \times V: U \in \mathcal{T}_x, V \in \mathcal{T}_y\right\}$(i.e. open sets in $X \times Y$ are formed of unions of sets of the form $U \times V$). It follows that if $(x, y) \in W \subset X \times Y$, with $W$ an open set in $X \times Y$, there exist $U \in \mathcal{T}_X$ and $V \in \mathcal{T}_Y$ such that $ (x, y) \in U \times V \subset W$ If $X\times Y$ is compact then $X$ and $Y$ are compact since the projections are continuous and [[Continuous Images of Compact Spaces are Compact]]. The converse is also true: > [!NOTE] Theorem \[MA260\] > If $\left(X, \mathcal{T}_X\right)$ and $\left(Y, \mathcal{T}_Y\right)$ are [[Compact topological spaces|compact topological spaces]] then $X \times Y$ is compact (with the product topology). **Proof**: Suppose that $\mathcal{U}$ is an open cover of $X \times Y$. We first show the following claim. The [[Tube Lemma]] asserts that for each $y \in Y$ there is an open set $N(y) \subset Y$ with $y \in N(y)$ such that $X \times N(y)$ can be covered by a finite subfamily of $\mathcal{U}$. The family $\{N(y): y \in Y\}$ forms an open cover of $Y$, so there is a finite subcover $\left\{N\left(y_1\right), \ldots, N\left(y_n\right)\right\}$ that covers $Y$. Thus $ X \times Y=\bigcup_{j=1}^n X \times N\left(y_j\right) $ This is a finite union, and, by the tube lemma, each of the sets $X \times N\left(y_j\right)$ can be covered by a finite subfamily of $\mathcal{U}$, so $X \times Y$ can be covered by a finite subfamily of $\mathcal{U}$. $\square$