# Let $A, B$ be closed subsets of $S^1$ such that $S^1 = A \cup B$. Prove that at least one of $A$ and $B$ contains a pair of mutually antipodal points.

Let $A, B$ be closed subsets of $S^1$ such that $S^1 = A \cup B$. Prove that at least one of $A$ and $B$ contains a pair of mutually antipodal points.

I am totally stuck on it.can I get some help

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This is probably too general for an answer, but may be of interest to you: the Lusternik-Schnirelmann theorem states that from a covering of $S^n$ by $n+1$ closed sets we can pick a set which contains antipodal points. (This is also true if you replace closed with open.) – tomasz May 8 '13 at 14:51

Since $S_1$ is connected, $A \cap B$ must be non-empty (or one of $A,B$ is $S^1$).

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As $S^1$ is connected, $A\cap B\ne \emptyset$. If $z\in A\cap B$, then $z$ and $-z$ are both in $A$ or both in $B$.

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Suppose $A$ and $B$ both do not contain a pair of mutually antipodal points. Then, if $x\in A$, we have $-x\notin A$ and so $-x\in B$. Hence $x\notin B$ and so, $A\cap B=\emptyset$. Given that $A$ is closed, $S^1\setminus A$ is open, but $S^1=A\cup B$ and $A\cap B=\emptyset$ so $S^1\setminus A=B$ implies that $B$ is open, which is a contradiction as non-trivial open proper subsets of $S^1$ can not be closed.

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Color the points that are only in $A$ red, color the points that are only in $B$ blue, color the points that are in $A$ and $B$ purple.

Hint: Take a diameter. If both points are the same color, we are done. Otherwise, show that there must be a purple point.

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