# Is the union of two circles in $R^{2}$ a locally Euclidean space?

Let $S$ be the topological space defined by the union of the two circles in $R^{2}$ with radius 1 and centers $(-1,0)$ and $(1,0)$, and the topology induced by the usual one of $R^{2}$.

Is $S$ a locally Euclidean Space?

I have got some ideas to prove that it is not a locally Euclidean Space, but I am stuck in the last steps.

My proof is:

Let $V\subset S$ be a neighborhood of $(0,0)$, and inside the unit open ball centered at the origin. If $S$ is locally Euclidean, then $V$ is homeomorphic to $R$, then $V$-{$(0,0)$} is homeomorphic to $R$-{$0$}.

Then, I tried to find a contradiction. I got some ideas from online sources which claim that the homeomorphism between $V$-{$(0,0)$} is homeomorphic to $R$-{$0$} will not happen since $V$-{$(0,0)$} has at least four connected components but $R$-{$0$} has only two.

I am really confused about the claim, and I checked the definition of connected component, but I still don't understand. How to prove that $V$-{$(0,0)$} has at least four connected components but $R$-{$0$} has only two? and how do you get the idea of the number of connected components?

Thank you so much!

• Your two circles are tangent. So they meet at a single point. Each circle itself is locally Euclidean. So removing the point of intersection says that each sufficiently small nbh will generate two connected components for each circle. Since the two circles only intersect at that point, there must be four components. – Daniel Feb 3 '18 at 4:01
• So are you asking for a rigorous proof of the fact that $V \setminus \{(0,0)\}$ has exactly four connected components, while $\mathbb R \setminus \{0\}$ has two connected components? Of course, your proof as it stands is correct. – астон вілла олоф мэллбэрг Feb 3 '18 at 4:02
• @Daniel Thank you! Make great sense to me. – JacobsonRadical Feb 3 '18 at 4:21
• @астонвіллаолофмэллбэрг Daniel does give me great sense of why that happens. Rigorous proof is absolutely great, but if it is not really convenient for you to write them all down (may be it will be really hard), it will be okay for me to not have the proof. Thank you for the confirmation of my proof! – JacobsonRadical Feb 3 '18 at 4:23
• @JacobsonRadical I have already written an answer, which I believe lends itself to more use than just this question. I wish to post it. The point is, the idea above is correct, however it's better to go through a rigorous treatment in case it is required in future.May I do so? – астон вілла олоф мэллбэрг Feb 3 '18 at 4:26

The idea, while not being enough to comprise a proof, is very simple : on paper, removing the point $0$ from $\mathbb R$ gives two parts which seem "connected" : one on either end of zero. $V \setminus \{(0,0)\}$ has four parts : to convince yourself, draw a small circle around the point $(0,0)$ and see that the removal of that point results in four different sections. Anyway, this was said in the comments as well, but for the completeness of the answer, it must be said.

Here is a nice trick to show that $\mathbb R \setminus \{0\}$ has exactly two connected components, while $V \setminus \{(0,0)\}$ has four connected components.

First, the rather not so interesting statement:

If two spaces are homeomorphic, there is a bijection between their connected components. In particular, if the number of connected components of $X$ is finite, the the number of connected components of $Y$ is equal, and therefore also finite.

Of course, since if $X$ and $Y$ are homeomorphic, then since $f$ and $f^{-1}$ are both continuous, they both map connected components to connected components, so from here you should be able to figure out the bijection.

Second :

A subset of $\mathbb R$ is connected if and only if it is an interval.

This is a fairly standard proof, I request you to look it up . It does not use the next result, but rather uses the g.l.b property, which characterizes the real line.

Third, the following amazing statement :

Let $X$ be a set. Let $Y_n = \{1,2,\ldots,n\}$ with the discrete topology i.e. every subset is open. Then, $X$ has atleast $n$ connected components if and only if there is a continuous surjective function from $X$ to $Y_n$.

In one direction, let $X$ have $m$ connected components and let $n \leq m$. Number all the connected components from $1$ to $n$ so that all numbers from $1$ to $n$ are covered, which can be done as $n \leq m$. Then, for $x \in X$, send $x$ to the number of the component which $x$ belongs to. This is surjective since every number is the image of some connected component. On the other hand it is continuous, since the inverse image of any subset of $Y$ is just a union of connected components, which is open. So we are done.

In the other direction, suppose that $X$ has less than $n$ connected components. Then, let $f$ be a continuous surjection from $X$ to $Y_n$. It is clear that $f$ takes connected sets to connected sets, so each connected component of $X$ must go to the same number, but then $X$ has less than $n$ connected components, so some number in $Y$ must miss out, which can't happen : $f$ is surjective.

Using the three above facts, now you have a rigorous treatment in front of you:

• Map $S= \mathbb R \setminus \{0\}$ to $\{1,2\}$ via $f(x) = 1$ if $x < 0$ and $2$ otherwise. This is continuous, and surjective, so $S$ has at least two connected components. But the two parts are intervals so they are already connected, so $S$ has exactly two connected components.

• Find the following description of $T = V \setminus \{(0,0)\}$ : without loss of generality we can assume that $V$ was actually a ball around the origin, of radius $r < 1$, so then $T$ has a nice description . You can find a constant $c$ depending on $r$ so that : $$T = \{(x_1,x_2) : |(x_1,x_2) - (1,0)| = 1 , -c < x_2 < 0\} \\ \quad \cup \{(x_1,x_2) : |(x_1,x_2) - (1,0)| = 1 , c > x_2 > 0\} \\\quad \cup \{(x_1,x_2) : |(x_1,x_2) - (-1,0)| = 1 , -c < x_2 < 0\} \\\quad \cup \{(x_1,x_2) : |(x_1,x_2) - (-1,0)| = 1 , c > x_2 > 0\}$$

• Drawing a diagram will convince you of the relationship between $c$ and $r$, but I intend not to derive it. The key point : the above sets are disjoint by triangle inequality, and furthermore, each one of them is homeomorphic to an interval, since we can parametrize the circle of radius one via the angle it makes with the $x$-axis or something like that, and then map each point in the set to the parameter it corresponds to.

• Finally, conclude that $V$ is homeomorphic to four disjoint intervals in $\mathbb R$. We can do what we did earlier : find a map to $\{1,2,3,4\}$ which is continuous and surjective, and note that there are four intervals to find that $T$ has four connected components.

So, there's a lot that goes on behind the scenes. I hope to have uncovered some of the magic lying behind these uncolourful statements.