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Let $X$ be the unit circle in $\mathbb{R}^2$; that is $X=\{(x,y):x^2+y^2=1\}$ and has the subspace topology. Consider $Y$ to be the subspace of $\mathbb{R}^2$ given by $Y=\{(x,y):x^2+y^2=1\}\cup\{(x,y):(x-2)^2+y^2=1\} $.

Is $Y$ homeomorphic to the space $X$?

I know I could use path-connectedness to prove $X$ and $Y$ are not homeomorphic but I am not supposed to. I thought of cardinality, but I really have no clue on how to approach this problem.

Question:

How should I prove $X$ and $Y$ are not homeomorphic?

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I don't know if this is what you have in mind whan you claim that you “could use path-connectedness to prove $X$ and $Y$ are not homeomorphic”; if it is, I will delete it.

If you remove a point from $X$, what remains is connected. But if you remove $(1,0)$ from $Y$, what remains becomes disconnected.

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  • $\begingroup$ The thing is that in this section of the book the path-connectedness definition has not been introduced yet. So I guess I could not infer that the circumferences union is disconnected without the theorem that implies that path-connectedness leads to connectedness . But thanks to the answer! $\endgroup$ – Pedro Gomes Oct 31 '18 at 18:22
  • $\begingroup$ @Pedro, but in this case it's easier to show that $Y$ minus the point is not connected directly from the definition of connectedness than using path connectedness. $\endgroup$ – Ennar Oct 31 '18 at 18:24
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    $\begingroup$ Then notice that I did not mention path-connectedness at all in my answer. Only connectedness. $\endgroup$ – José Carlos Santos Oct 31 '18 at 18:26
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    $\begingroup$ That's easy: the left half of $Y$ is clopen. $\endgroup$ – José Carlos Santos Oct 31 '18 at 18:28
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    $\begingroup$ The left side of $Y$ is closed becaue it is the intersection of $Y$ with the closed halfplane $\{(x,y)\in\mathbb{R}^2\,|\,x\leqslant0\}$ and it is open because it is the intersection of $Y$ with the open halfplane $\{(x,y)\in\mathbb{R}^2\,|\,x<0\}$. $\endgroup$ – José Carlos Santos Oct 31 '18 at 18:32

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