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Given $p: \mathbb{R} \times \mathbb{R} \to \mathbb{R}, (x,y) \mapsto x$

Why is $p$ open but not closed?

Shouldn't it be the case $p$ is closed not open because it sends a singleton to singleton, and all singletons in Euclidean spaces are closed?

Fraïssé
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    There are many closed sets that are not singletons. – Hoot Jun 05 '16 at 03:27
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    You seem to have made the same error of trying to prove a statement of the form "For all closed $A$, ..." by considering only singletons in this other question of yours. It's perhaps worth watching out for the error of treating on a specific instance of "for all" when it doesn't represent the general case. – Milo Brandt Jun 05 '16 at 03:31
  • Please don’t take it amiss when I say that we should never ask why something isn’t true. What we need in all cases is a proof when it is true. – Lubin Jun 05 '16 at 03:48

2 Answers2

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The set $F=\{(\frac{1}{n},n):n\in\mathbb{N}\}$ is a closed subset of $\mathbb{R}^2$ because it has no limit points, but its projection onto the $x$ coordinate is the set $\{\frac{1}{n}:n\in\mathbb{N}\}$, which is not closed because it doesn't contain the limit point $0$. Therefore the projection is not a closed map.

carmichael561
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  • Would have never thought of that example – Fraïssé Jun 05 '16 at 03:42
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    @Lookbehindyou Would you have thought of the example ${(x,y):xy=1}$? It's a closed set (because it's a level set of the continuous function $f(x,y)=xy$) but its projection $\mathbb R\setminus{0}$ is not closed. – bof Jun 05 '16 at 03:47
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You could also think about the set $S=\{ (t, \tan(t)): -\pi/2<t<\pi/2\}. $

This set is closed in $\mathbb{R}^2$, but its projection over the $x$-axis is not closed in $\mathbb{R}$.

rowcol
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