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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?

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    $\begingroup$ The only point whose removal disconnects $Y$ is $\langle 1,0\rangle$, but the removal of any non-endpoint of an interval disconnects the interval. A homeomorphism takes cut points to cut points, so there is no homeomorphism from $Y$ to an interval. $\endgroup$ – Brian M. Scott Jan 18 at 7:49
  • $\begingroup$ Just to make sure I am applying the theorem correctly, it states that If X and Y are homeomorphic, there is also a homeomorphism if we remove any point from X and any point from Y right? So we can choose points to remove such that X and Y have different properties such as connectedness which means they are not homeomorphic right? $\endgroup$ – William Jan 18 at 7:57
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    $\begingroup$ Not quite. The point is that if $X$ and $Y$ are spaces, $h:X\to Y$ is a homeomorphism, and $x$ is a cut point of $X$, then $h(x)$ must be a cut point of $Y$, because $X\setminus\{x\}$ is homeomorphic to $Y\setminus\{h(x)\}$. This means that $h$ must be (among other things) a bijection between the cut points of $X$ and the cut points of $Y$. Your space $Y$ has only one cut point, while an interval has infinitely many, so there cannot be a bijection between the cut points of $Y$ and the cut points of an interval, and therefore there can be no homeomorphism between them. $\endgroup$ – Brian M. Scott Jan 18 at 8:17
  • $\begingroup$ oh ok got it! Thanks $\endgroup$ – William Jan 18 at 8:26
  • $\begingroup$ You’re welcome! $\endgroup$ – Brian M. Scott Jan 18 at 8:26
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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.

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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 $S^1$.

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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 $E'$.

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