MHB Projection Map $X \times Y$: Closure Property

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The discussion focuses on proving that the projection map \( p_X : X \times Y \to X \) is closed when \( Y \) is a compact topological space. Participants note that no solutions were provided for the problem of the week. The original poster shares their own solution, which likely involves demonstrating that the image of a closed set under the projection map remains closed in \( X \). The closure property of the projection map is tied to the compactness of \( Y \). Overall, the thread emphasizes the importance of understanding projection maps in topology.
Euge
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Here is this week's POTW:

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Let $X$ and $Y$ be topological spaces. If $Y$ is compact, show that the projection map $p_X : X \times Y \to X$ is closed.
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No one solved this problem. You can read my solution below.

Let $C$ be a closed set in $X \times Y$. If $x\in X\setminus p_X(C)$, then for every $y\in Y$ the ordered pair $(x,y)\notin C$; it follows that there are open neighborhoods $U_y$ of $x$ and $V_y$ of $y$ such that $(U_y\times V_y)\cap C = \emptyset$. The collection $\{V_y:y\in Y\}$ is an open cover of $Y$; by compactness of $Y$, there are $y_1,\ldots, y_n\in Y$ such that $Y = V_{y_1}\cap \cdots \cap V_{y_n}$. Let $U = U_{y_1}\cap \cdots \cap U_{y_n}$. Then $U$ is an open neighborhood of $x$ such that $(U\times Y) \cap C = \emptyset$, i.e., $p_X^{-1}(U)\cap C = \emptyset$. Thus $U \cap p_X(C) = \emptyset$. Since $x$ was arbitrary, $p_X(C)$ is closed.
 

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