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For every $A \subset \mathbb R^3$ we define $\mathcal{P}(A)\subset \mathbb R^2$ by $$ \mathcal{P}(A) := \{ (x,y) \mid \exists_{z \in \mathbb R}:(x,y,z) \in A \} \,. $$ Prove or disprove that $A$ closed $\implies$ $\mathcal{P}(A)$ closed and $A$ compact $\implies$ $\mathcal{P}(A)$ compact.

This $\mathcal{P}(A)$ seems to me the projection on the $xy$-plane, so intuitively both make sense to me.

My try

For the first one, to prove something is closed seems easiest to do with sequences, so for all sequences $(x^{(n)})$ in $A$ with $x^{(n)} \to x$ we have that $x \in A$. But I do not know how to progress further in a useful direction. For the second one I don't even know where to start.

Update

For the first one I tried to argue that for any sequence going to $x \in A$, a projection of that sequence on the $xy$-plane will also go to the projection of $x$, so $x \in \mathcal{P}(A)$, so hence $\mathcal{P}(A)$ must be closed.

PHPirate
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  • Can you show that $\mathcal{P}$ is continuous? Because then it follows that $A$ compact implies $\mathcal{P}(A)$ is compact (since the image under a continuous map of a compact set is always compact). – Chill2Macht Jun 26 '16 at 15:42
  • The question is essentially answered at http://math.stackexchange.com/questions/22697/projection-map-being-a-closed-map – guestDiego Jun 26 '16 at 15:51
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    @William Maybe I could try, but I would need to prove the theorem you suggest I think. I'll have a look at that – PHPirate Jun 26 '16 at 15:55
  • @guestDiego comment noticed, I'm still busy trying to translate it to my problem :) – PHPirate Jun 26 '16 at 15:56

3 Answers3

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Let $A=\{(x,0,z): xz=1\},$ which is closed in $R^3.$ Then $P(A)=\{(x,0):x\ne 0\},$ which is not closed in $R^2.$

Observe that $P$ is Lipschitz-continuous, as $\|P(u)-P(v)\|=\|P(u-v)\|\leq \|u-v\|.$ The continuous image of a compact space is compact. So if $A\subset R^3$ is compact then $P(A)$ is compact.

DanielWainfleet
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  • Your counterexample coincided with me finding out the crux between these two questions, indeed, as guestDiego mentioned in the other answer, the boundedness in the third component makes this second one true. Thanks for helping with Williams suggestion as well! (turned out we did prove that theorem earlier...) – PHPirate Jun 26 '16 at 16:19
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Hints:

  1. The fact that you are unable to prove this might suggest that the statement is actually untrue. Try and work out where your proof is falling apart and use that to construct a closed set $A\subset \mathbb R^3$ such that $P(A)$ is not closed.

  2. This is not as difficult as you're making it out to be. What definition of compactness are you using?

John Gowers
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  • Really? My intuition says that if you have any sequence going to $x \in A$, then a projection of that sequence on the $xy$-plane will also go to the projection of $x$, so $x\in \mathcal{P}(A)$... For the second, we defined compact as 'every covering has a finite subcovering', but also we know the Heine-Borel theorem. The problem seems similar to the first one to me... – PHPirate Jun 26 '16 at 15:52
  • The answer to the question is at http://math.stackexchange.com/questions/22697/projection-map-being-a-closed-map – guestDiego Jun 26 '16 at 15:55
  • @guestDiego $Y$ isn't compact here though – Chill2Macht Jun 26 '16 at 15:56
  • I updated the question, even though I don't know how to formally argue about it, I don't see any holes in my try? – PHPirate Jun 26 '16 at 16:04
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    Aaah wait a second, I think I see something. A set which is a function of two variables where $z$ goes to inifinity as $(x,y)$ nears some point could be a nice counterexample! Now 2. – PHPirate Jun 26 '16 at 16:13
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    Compactness is necessary for having the closedness of $A$. $Y$ can be chosen compact if $A$ is bounded in the third component, If it is not, the statement is not true in general – guestDiego Jun 26 '16 at 16:14
  • @PHPirate Well done. – John Gowers Jun 27 '16 at 15:59
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Hint:

The first one is not true.

For the second one try this:

I assume you know Heine Borel's theorem about compact sets.

Heine Borel says $A$ is bounded. So for all components of $(x_1,x_2,x_3)\in A $ you can find an $M>0$: $|x_i|<M$. That means $|x_i|^2<M^2$. Now we have: \begin{align} ||(x_1,x_2)||^2=|x_1|^2+|x_2|^2<2M^2\end{align}

Can you finish now?

Heine Borel says $A$ is also closed. Choose a convergent sequence $x^n \rightarrow x$ in $\mathcal{P}(A)$. There is a sequence $x^n_3$ with $(x_1^n,x_2^n,x_3^n)\in A$. You know $x_3^n$ is boundend. By Bolzano Weierstrass theorem for sequence we have now there is $x^{n_k}_3 \rightarrow x_3$. What can you say about the original sequence in $\mathcal{P}(A)$ now?

Combine them both and by Heine Borel you have $\mathcal{P}(A)$ is compact.

Shashi
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