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Consider $\mathbb{R}^2$ as $\mathbb{R} \times \mathbb{R}$ with the product topology. I am simply trying to show that the two projections $p_1$ and $p_2$ onto the first and second factor space respectively are not closed mappings. It seems like this should be easy, but I have not been able to come up with a closed set in $\mathbb{R}^2$ whose projection onto one of the axes is not closed.

I don't really have any work to show...I've really just tried the obvious things like closed rectangles and unions of such, the complement of an open rectangle or union of open rectangles, horizontal and vertical lines, unions of singletons, etc., and haven't come up with anything non-obvious, which I hope is where the answer lies. It's bothering me that I can't come up with an answer, and I'd appreciate some help. Thanks.

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    $\begingroup$ Take an hyperbole. $\endgroup$
    – Tomás
    Oct 26, 2012 at 23:24
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    $\begingroup$ @Tomás: The term hyperbole is actually from literature, and it means "exaggeration" (e.g. the pub was crowded, there were a million people there!); you probably wanted to say hyperbola which is a mathematical object. $\endgroup$
    – Asaf Karagila ♦
    Oct 27, 2012 at 1:38
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    $\begingroup$ Yes @AsafKaragila, hahha, thank you, You help me to improve my english (it is so bad haha) $\endgroup$
    – Tomás
    Oct 27, 2012 at 10:24

3 Answers 3

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No bounded set will work, because a closed and bounded set in ${\bf R}^2$ is compact, and the image of a compact set under any continuous map is compact (so closed in any Hausdorff space, in particular in $\bf R$).

On the other hand, the graph of any function with a vertical asymptote will work, for instance that of $1/x$.

In fact, it is not hard to show that any open set in $\bf R$ can be obtained as the projection of a closed set in ${\bf R}^2$ in such a way that the projection is injective (no two points in the closed set project onto the same point in $\bf R$), by a similar technique. This is related to the classical fact that any $G_\delta$ (countable intersection of open sets) in ${\bf R}$ (or any other Polish space, that is, separable and completely metrizable, if you're familiar with the concepts) can be embedded as a closed set in $\bf R^N$ (product of countably infinitely many copies of $\bf R$).

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    $\begingroup$ Thanks for making what could have been a very simple answer more significant, interesting, and even an introduction to higher concepts! $\endgroup$
    – Alex Petzke
    Oct 27, 2012 at 2:27
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    $\begingroup$ Someone downvoted this answer. Why? $\endgroup$
    – tomasz
    May 17, 2013 at 19:56
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The graph of $\tan^{-1}(x)$ is closed, but its projection onto the $y$-axis is not.

There's a related discussion in SO at Projection map being a closed map.

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  • $\begingroup$ I learned this example recently from @NateEldredge in [another SO question] (math.stackexchange.com/questions/214597/…). $\endgroup$
    – Hew Wolff
    Oct 26, 2012 at 23:26
  • $\begingroup$ Projection of the graph of $\tan (x)$ onto the $x$-axis is also not closed (and no wonder, it being symmetrical to the graph of $\tan ^{-1}(x)$). $\endgroup$
    – tomasz
    Oct 26, 2012 at 23:33
  • $\begingroup$ A slick counterexample indeed! $\endgroup$
    – Cheng
    Feb 7, 2021 at 11:29
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Consider the map $\phi : \mathbb{R}^2 \to \mathbb{R}$ that takes: $$(x,y) \mapsto x \cdot y.$$

This map is continous, thus $\phi^{-1}(1)$ is closed in $\mathbb{R}^2$. On the other hand its projection onto the first coordinate is $\mathbb{R} -\{0\}$ which is not closed because it's open and $\mathbb{R}$ is connected.

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