Show that if $Y$ is compact, then the projection $\pi_{X} : X \times Y \rightarrow X$ is a closed function.
I can't to solve the question above. Indeed, I think that projection is a closed (and opened) function even without Y compact. I'm right?
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2You definitely need $Y$ compact. Think about the graph of $y=1/x$ projected onto the first coordinate in $\mathbb R^2$. – Cheerful Parsnip Jun 28 '13 at 03:37
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Projection map is always an open map. but for being a closed map $Y$ has to be compact. Take the graph of $y=1/x$ in first quadrant. It is closed. But its projection onto the $X$- axis is positive reals which is not closed. – tessellation Jun 28 '13 at 03:39