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I am trying to answer the question stated in the title.

The hint in my book says to realize that for any z on the circle C{z} is still connected.

I believe I can deal with case that shows that C{z} is not homeomorphic to any circle, but I am not sure how to generalize this to any subset of C.

We note that C is homeomorphic to any circle because any circle is just a dilation and a translation of C and dilations and translations are homeomorphic. Thus, the problem is reduced to showing that C{z} is not homeomorphic to C.

Well, if there did exist a homeomorphism from C{z} to C, then it would have to take connected components to connected components since homeomorphisms are continuous by definition.

So, with a rotation composed with our homeomorphism, we should be able to fix a point w in C{z} so that our rotation with our homeomorphism, call it g, satisfies g(w) = w. But then we should get that the that the inverse of g at z, g^(-1)(z) is undefined. This contradicts that g is a homeomorphism.

Can anyone tell me if I'm on the right track and how to generalize this to any subset of C?

Thanks

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Hint since you're overthinking it: If $X$ is a proper subset of the circle, then there is at least one $z\in C$ that is not an element of $X$, so $X\subseteq C-\{z\}$. The space $C-\{z\}$ is homeomorphic to an interval $I$ in $\mathbb{R}$, so identify $X$ with a subset of $I$.

Now what do you know about removing points on intervals and connectedness? How does this tie in with the hint you were given?

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    $\begingroup$ If I is our interval and if we remove points from it, then I will separated into multiple components. Call the union of these multiple components I'. If I' is homeomorphic to I then it take connected components to connected components. Thus, a homeomorphism would have to take I to one of the connected components and not to any other components. Thus, no homeomorphism exists. Any subset of C is homeomorphic to one of the intervals with multiple components; thus any subset of C cannot be homeomorphic to C. Thank you Neal. Is this what you had in mind? $\endgroup$
    – Clyde
    Commented Mar 15, 2014 at 3:04

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