Let $\pi: X \times Y \to X$ be the projection map where $Y$ is compact. Prove that $\pi$ is a closed map.
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First I would like to see a proof of this claim.
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I want to know that here why compactness is necessary or do we have any other weaker condition other than compactness for the same result to hold.
Best Answer
Suppose $Z \subset X \times Y$ is closed, and suppose $x_0 \in X \setminus \pi[Z]$. For any $y \in Y, (x_0, y) \notin Z$, and as $Z$ is closed we find a basic open subset $U(y) \times V(y)$ of $X \times Y$ that contains $(x_0, y)$ and misses $Z$. The $V(y)$ cover $Y$, so finitely many of them cover $Y$ by compactness, say $V(y_1),\ldots,V(y_n)$ do. Now define $U = \cap_{i=1}^{n} U(y_i)$, and note that $U$ is an open neighbourhood of $x_0$ that misses $\pi[Z]$: suppose that there is some $(x,y)\in Z$ with $\pi(x,y) = x \in U$. Then $y \in V(y_i)$ for some $i$, and as $x \in U \subseteq U(y_i)$ (as $U$ is the intersection of all $U(y_i)$) we get that $(x,y) \in (U(y_i) \times V(y_i)) \cap Z$ which contradicts how these sets were chosen to be disjoint from $Z$. So $U \cap \pi[Z]=\emptyset$ and $\pi[Z]$ is closed.
To see that the closed projection property implies compactness: suppose $X$ has the closed projection property along $X$, and let $\cal{F}$ be a filter on $X$. Define a space $Y$ that is as a set $X \cup \{\ast\}, \ast \notin X$, where $X$ has the discrete topology and a neighbourhood of $\ast$ is of the form $A \cup \{\ast\}$ with $A \in \cal{F}$. Then $D = \{(x,x): x \in X\}$ is a subset $X \times Y$ and closedness of the projection $p: X \times Y \rightarrow Y$ implies that some point $(x,\ast)$ is in its closure: $D$ cannot be closed in $X \times Y$, because $p[D] = X$ is not closed in $Y$, as $\ast \in \overline{p[D]} = p[\overline{D}]$, by closedness (and continuity) of $p$. This $x$ is an adherence point of the filter, because if $A \in \mathcal{F}$ and $x \in O$ where $O$ is open in $X$, then $O \times (A \cup \{\ast\})$ is basic open in $X \times Y$ and so intersects $D$ in some $(p,p)$, $p \in X$. This $p \in O \cap A \neq \emptyset$, showing that $x$ is an adherence point of $\mathcal{F}$.