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Aug 27, 2012 at 10:26 comment added Ilya Bogdanov The correct statement seems to be as follows. Assume that the intersection of any $t+1$ of $C_i$'s is empty. Then there exists a connected component of $S\setminus \cup C_i$ which touches every $t$-dimensional face of $S$.
Aug 27, 2012 at 9:33 vote accept domotorp
Aug 26, 2012 at 16:36 comment added Ilya Bogdanov Well, the reasoning from below shows that if $C_1,\dots,C_t$ do not share a common point, then there exists a small (full-dimensional) ball attached to $n+1-t$ hyperfaces. But you are right, it seems that something better holds...
Aug 26, 2012 at 15:55 comment added domotorp As I defined linear subspace of S, it is required that it passes through the origin if we extend it beyond S. I am sorry that I defined it so vaguely that it is hard to understand what I want. Anyhow, now it is unimportant, as your example showed that we cannot have it.. Do you have any ideas about some theorems that could guarantee some high dim manifold stretching to many boundaries of S?
Aug 26, 2012 at 8:11 comment added Ilya Bogdanov Now I don't understand. What is the principal difference between affine and linear subspace, if (as in your definition) $S$ does not contain the origin?
Aug 26, 2012 at 7:10 history edited domotorp CC BY-SA 3.0
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Aug 26, 2012 at 6:44 history edited domotorp CC BY-SA 3.0
fixed def of subspace of S
Aug 25, 2012 at 20:58 comment added domotorp Why does it become false?
Aug 25, 2012 at 20:48 comment added Ilya Bogdanov In view of the answer below, I wish to change the notion of $k$-dimensional affine subspace; it seems that this subspace should be `large enough'. On the other hand, if it will be too large, then the conjecture becomes false...
Aug 25, 2012 at 20:10 answer added Ilya Bogdanov timeline score: 1
Aug 25, 2012 at 18:06 history edited domotorp CC BY-SA 3.0
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Aug 25, 2012 at 16:04 history asked domotorp CC BY-SA 3.0