Tube lemma: Consider the product space , where Y is compact. If N is an open set containing the slice of , then N contains some tube about , where V is a neighborhood of in .

We shall show that is open in X using a tube lemma, which is equivalent to showing that is closed in X.

Let be any element in . By the tube lemma, there exists a tube such that

.

Now, for any in , we have an open set V containing such that . Thus, is open in .