# 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?

• 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. – Brian M. Scott Jan 18 at 7:49
• 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? – William Jan 18 at 7:57
• 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. – Brian M. Scott Jan 18 at 8:17
• oh ok got it! Thanks – William Jan 18 at 8:26
• You’re welcome! – Brian M. Scott Jan 18 at 8:26

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.

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 $$S^1$$.

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 $$E'$$.