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I am reading the proof of that the product of finitely many compact spaces is compact from munkres their is a certain little step I don't understand 100 %.

The way he proved it is that he divided the proof into two steps:

Step 1: Let N is an open set around the slice $x_0 \times Y$ he proved that their is a tube i.e open set W in X that contains $x_0$ such that N contains $W\times Y$.

I understand the proof for step 1, however their is little part I don't understand in step 2.

Step 2: Let X and Y be compact spaces. Let $A$ be a open covering of $X\times Y$. Given $x_0$, the slice $x_0 \times Y$ is compact since it is homeomorphic to Y, so it can be covered by finitely many elements $A_1,...,A_m$ of A. Their union $N = A_1 \cup ... \cup A_m$ is an open set containing $x_0 \times Y$. Here is my question why does it contain $x_0 \times Y$ shouldn't their union be equal to $x_0 \times Y$ by the definition of compactness ? because this will make a huge difference in the proof in the other parts.

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For a compact space you are right: The famous finite subcover is a collection of open subsets of the space such that theri union equals the space. Here we talk about a compact subspace, so we have to deal with relative topology, or relatively open sets. By definition of that, the $A_i\cap (\{x_0\}\times Y)$ are open in the space $\{x_0\}\times Y$ (or which is almost the same, in $Y$). That the union of finitely many $A_i\cap (\{x_0\}\times Y)$ equals $\{x_0\}\times Y$ is just another way of saying that the union of the $A_i$ contains $\{x_0\}\times Y$.

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The sets $A_i$ come from the original covering $A$ which consists of subsets of $X\times Y,$ not necessarily contained in $x_0\times Y.$

If the definition of compactness that you are using requires open subsets of the set under scrutiny, consider the intersections $A_i\cap(x_0\times Y).$

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As a sanity check, $x_0 \times Y$ is not necessarily open in $X \times Y$, so the union is not necessarily $x_0 \times Y$. What you might be thinking is that the union of the sets $A_i \cap (x_0 \times Y)$ will indeed be $x_0 \times Y$.

A picture you might want to have in mind is $X \times Y$ being a rectangle (e.g. in $\mathbb{R} \times \mathbb{R}$), and the slice $x_0 \times Y$ being a vertical line. The cover could be open balls covering $X \times Y$. The specified union is a finite union of balls covering the vertical line $x_0 \times Y$, and the union need not be $x_0 \times Y$, although intersecting each open ball with $x_0 \times Y$ before taking the union will indeed yield exactly $x_0 \times Y$.

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