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Let $X,Y\subset \mathbb{R}^n$ are compact and convex. Define $S=\{\langle x,y\rangle\in \mathbb{R}\mid x\in X, y\in Y\} $. Is $S$ convex?

I tried to prove by taking an arbitrary convex combination of two elements in $S$ and tried to show it is also in $S$ but I got nowhere. I don't even know if the statement is true. Is it true? If so, any hint on how to prove it would be greatly appreciated.

daw
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curiosity
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    The question is phrased unnecessarily complicated because the convex sets within $\mathbb R$ are specifically the intervals $(a,b), [a,b), (a,b], [a,b]$. – Wolfgang Bangerth Feb 08 '24 at 22:15
  • @WolfgangBangerth: ...where $a$ might be$-\infty$ and $b$ might be $+\infty$. – TonyK Feb 09 '24 at 00:30

1 Answers1

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It is true, and it suffices that $X,Y\subset \mathbb{R}^n$ are connected. (This is more general since all convex sets in $\Bbb R^n$ are connected.)

Then $X \times Y$ is also connected (see e.g. here) and the image of $X \times Y$ under the continuous function $f(x, y) = \langle x,y\rangle$ is a connected subset of $\Bbb R$ (see e.g. here).

The connected subsets of $\Bbb R$ are exactly the intervals, and intervals are convex sets.

If, in addition, both $X$ and $Y$ are compact then $S$ is a compact interval, i.e. $S = [a, b]$ with real number $a \le b$.

Martin R
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  • Convex sets in $\mathbb R^n$ are always connected, aren't they? – nelynx Feb 08 '24 at 13:53
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    @nelynx: Yes, but connected sets are not necessarily convex. What I wanted to express is that the condition “Let $X,Y\subset \mathbb{R}^n$ are compact and convex” can be relaxed. – Martin R Feb 08 '24 at 13:54
  • I would add that to the answer (since the question originally asks for convex sets), but yeah, that is the best argument – nelynx Feb 08 '24 at 13:58
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    @MartinR Beautiful! It is a world different from the brute force approach I was trying. Your answer just made my day, thank you so much. – curiosity Feb 08 '24 at 15:26