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I am sorry to ask this trivial question:

Let $R$ be a square $[0,1]\times[0,1]$, and $A$ is a continuous curve from $(0,0)$ to $(1,1)$, while $B$ is another continuous curve from $(0,1)$ to $(1,0)$. Show that $A$ and $B$ always intersect.

I have tried several methods, e.g., fixed point theorem and planar graph embedded on a Möbius band and etc. And I think this question is related to compactness and Hausdorff.

Thanks.

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    $\begingroup$ This is a classic problem. In this thread on MathOverflow you'll find a long list of different approaches. $\endgroup$
    – t.b.
    Nov 2, 2011 at 4:56

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You do need to specify that both your curves are required to remain in the square, otherwise one curve could go outside (draw a picture)...

That being said, if both curves remain inside the square, what goes wrong if we draw a fifth point P outside the square, then draw continuous curves from P to each of the four corners of the square, all the four curves staying outside the square (and not touching each other) until they reach the corners? If your two original curves somehow do not intersect, what has been drawn? Again, draw some pictures.

EDIT: you used the words planar graph, that should be enough of a hint. Forget the Moebius band.

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  • $\begingroup$ Quite a nice proof. Thanks very much! $\endgroup$
    – user18705
    Nov 3, 2011 at 1:47

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