Consider compact set $\Omega\subset\mathbb{R}^d$, whose intersection with any $(d-1)$-dimensional subspace of $\mathbb{R}^d$ is contractible. Then, is such $\Omega$ convex?
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@amsmath - yes, to be exact it is contractible. I edited the question. sorry again. – Moonshine Jun 05 '19 at 14:02
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This is trivially true for $d=2$ as contractible subsets of $\mathbb R^1$ are intervals. – daw Jun 07 '19 at 11:39
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No. Consider a large sphere, with a smaller sphere removed from the centre. The intersection with any plane is still connected (it will be a circle or an annulus). But this is clearly not a convex set. Similar higher-dimensional analogues can be found.
auscrypt
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I should have added condition of contractibility, I forgot to write that. Good observation and sorry for missing that. – Moonshine Jun 05 '19 at 14:05
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