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For an application in optimization, I came across the following problem:

(Notation: Let $[0,1]^N$ denote a real hypercube. For a vector $t\in [0,1]^N$, let $\hat{t}_i\in [0,1]^{N-1}$ denote the removal of the $i$-th coordinate.)

Let $g_i:[0,1]^{N-1}\to [0,1]$ be continuous functions and define hypersurfaces \begin{align} S_i:=\{t\in [0,1]^N: t_i=g_i(\hat{t}_i)\}. \end{align}

Is the intersection \begin{align} S_1\cap \cdots \cap S_N \end{align} necessarily non-empty?

This is a generalized version of intersection of continuous curves in a square., where the answer is ''yes''.

For my application, it is not a restriction to assume that $g_i$ are smooth.

Rob Arthan
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Cellardoor
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  • @RobArthan, in this case the sets ${(t_1,c_1):t_1\in [0,1]}$ and ${(c_2,t_2):t_2\in [0,1]}$ are straight lines in the square $[0,1]^2$. Two such straight lines clearly meet in a point. – Cellardoor Nov 22 '23 at 22:26
  • @Carlyle: If we put $N=2$, then we get a square $[0,1]^2$. Our two hypersurfaces $S_1,S_2$ are curves in this square (they are graphs defined by equations $t_2=g_2(t_1),t_1=g_1(t_2) $ for some continuous $g_1,g_2$. For instance, $S_1={(g_1(t_2),t_2): t_2\in [0,1]}$). $S_1$ is a continuous curve meeting the lower and top edges of the square, and $S_2$ is a continuous curve meeting the left and right edges of the square (or the other way around.) Does this clarify? – Cellardoor Nov 22 '23 at 22:28
  • @Cellardoor, yes thank you, an interesting question – Carlyle Nov 22 '23 at 22:43
  • Makes sense now. I'll delete my comments. – Rob Arthan Nov 22 '23 at 22:44

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