Is the union of two circles homeomorphic to an interval?

Question

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?

Answer 1

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.

Answer 2

Here's an alternative argument that covers wide range of spaces, regardless of cut points.

Let $X$, $Y$ be any topological spaces. Consider a homeomorphism $f:X\to Y$. Now let $Z$ be any topological space and $\alpha:Z\to X$ be a continuous function. Then $\alpha$ is injective if and only if $f\circ\alpha$ is injective. Which is easy to see by applying $f$ to $\alpha$ and $f^{-1}$ to $f\circ\alpha$.

In particular $X$ admits an injective map $Z\to X$ if and only if $Y$ admits an injective map $Z\to Y$.

This shows that your $Y$ (or more generally any space containing $S^1$ as a subspace) cannot be homeomorphic to the interval $[0,1]$. Because there is an obvious injective continuous map $S^1\to Y$ while there is no continuous injective map $S^1\to [0,1]$. That's because every map $S^1\to [0,1]$ arises from a map $[0,1]\to [0,1]$ with the same values at endpoints and so by the intermediate value property it cannot be injective on $(0,1)$, which then "lifts" to $S^1$.

Answer 3

Just for a change: Let $D$ be the open disk in $\Bbb R^2$ centered at $p=(1,0)$ with radius $1$ and let $E=D\cap Y.$ Then $E$ is a connected subspace of $Y$ and $E\setminus \{p\}$ is the union of $4$ pairwise-disjoint non-empty connected open subspaces of $E$. But if $E'$ is any connected subspace of $\Bbb R$ and $p'\in E'$ then $E'\setminus \{p'\}$ is either connected or is the union of $2$ disjoint non-empty connected open subspaces of $E'$.