Is the union of two circles homeomorphic to an interval? Let $Y$ be the subspace of $\Bbb R^2$ given by $Y=\{(x,y): x^2+ y^2=1\}\cup \{(x,y): (x−2)^2+ y^2=1\}$. Is $Y$ homeomorphic to an interval?
I have previously already shown that the unit circle is not homeomorphic to any interval and I think this is also true for $Y$. Basically if we remove a point from $Y$ that is not the intersection, and remove a point from any interval that is not an endpoint, the interval becomes disconnected, but the union of two circles are still connected so there is no homeomorphisms. Is that correct?

A cutpoint of a connected space $X$ is a $p \in X$ such that $X\setminus\{p\}$ is disconnected.
If $f:X \to Y$ is a homeomorphism of connected spaces $X$ and $Y$ and $p$ is a cut point of $X$ then $f(p)$ is a cutpoint of $Y$ (and vice versa).
If we take $X$ to be an interval, then $X$ has at most two non-cutpoints. (the endpoints in the case of a closed interval). $Y$ on the other hand has infinitely many non-cutpoints (all points except $(1,0)$). So there can be no homeomorphism between them by the observations in the second paragraph.
