# Let C be a circle. Show that the only subset of C homeomorphic to a circle is C itself

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

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 $I$.