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I know that the unit circle = $\{(x,y): x^2+y^2 =1\}$ is not homeomorphic to the closed unit disk = $\{(x,y): x^2+y^2 \leq 1\}$, but I'm not sure how to prove it. I've tried with arguments with cut-points and with (path)connectedness, but still not getting a good argument. Any help?

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If you remove two points from the circle, it becomes disconnected.

The disk on the other hand remains connected after such removal. To see the latter observe that there are infintely many pairwise disjoint (apart from the ends) paths between any two points of the disk, and removing finitely many poitns can only destroy finitely many of them.

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The unit disk $B=\{x^2+y^2\leq1\}$ has fundamental group $\pi_1(B)=\{0\}$, while the circle $S^1=\{x^2+y^2=1\}$ has fundamental group $\pi_1(S^1)=\Bbb Z$. And a necessary condition for two topological spaces to be homeomorphic is that they have the same fundamental group (or better isomorphic groups).

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Any open cover of the circle has a refinement such that the intersection of any three distinct sets is empty. But there is an open cover of the disc that admits no such refinement:

enter image description here

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