Let $N$=$\{1+\frac 1 n\;|\;n\in\mathbb N\}\cup\{1\}$. I'm interested in whether the space $$ A=\cup_{n\in N}A_n \text{ where } A_n:=\{(x,y)\in\mathbb R^2\;|\;(x-n)^2 + y^2 = n^2, x\leq 1\} $$ is a retract of $\mathbb R^2$. Note that since $1\in N$, the limiting circle is also in $A$. The space is the red part of the following picture:
First, observe that the space is path connected, since all circles intersect in $(0,0)$. Secondly, the space is closed, because it is an intersection of a descending chain of closed sets (or, if you prefer, because all of its accumulation points are its elements). Since we fail to conclude with both of these arguments that $A$ is not a retract of $\mathbb R^2$, we next try to prove that it is a retract.
I've tried these things, none of which worked out:
- My first idea was to try and combine "croissants" (you'll see what I mean if you look at the picture) into the disk $\mathbb D((2,0),2)$. It's possible to show that for $n\geq 2$ $A_n\cup A_{n+1}$ is a retract of the croissant $C_n=\{(x,y)\in\mathbb R^2\;|\;(x-n)^2 + y^2 \leq n^2\text{ and }(x-n\frac{n-1}{n+1})^2 + y^2 \geq (n\frac{n-1}{n+1})^2\}$ and also that $A_1$ is a retract of $\mathbb D((1,0), 1)$. (I won't go into detail of how to do it as it is not so relevant.) When I tried to combine the retractions piecewisely on the closed cover $\{C_n\;|\;n\in N\setminus\{1\}\}\cup \{\mathbb D((1,0),1)\}$ of $\mathbb D((2,0),2)$ into a continuous map $r\colon\mathbb D((2,0),2)\rightarrow A$, I encountered two problems. Firstly, there was no reason to think that the retraction maps $r_n\colon C_n\rightarrow A_n\cup A_{n+1}$ are consistent on their intersections, so the combined map isn't even well-defined. Secondly, $(0,0)$ is problematic; since there is no neighborhood of $(0,0)$ that would intersect only finitely many sets in the closed cover (i.e. the given cover is not locally finite), we cannot guarantee that the piecewisely defined map $r$ is continuous. This second issue essentially means that if you try to come up with a piecewisely defined retraction of some subspace of $\mathbb R^2$ to $A$, you will have problems with continuity, because the point $(0,0)$ doesn't have a neighborhood that would intersect finitely many $A_n$'s. I also tried to combine retractions of the circles $S_n=\{(x,y)\in\mathbb R^2\;|\;(x-n)^2 + y^2 = n^2\}$ to $A_n$ to yield a retraction of $\cup_{n\in N}S_n$ to A (if that could, in a separate attempt, somehow help us to get a retraction of $\mathbb R^2$ to $A$) and encountered the same issue.
- So, I began to suspect that $A$ is not a retract of $\mathbb R^2$. Since we know that every retract of an absolute extensor is itself an absolute extensor, we could somehow use it to prove with a contradiction that $A$ is not a retract of $\mathbb R^2$ (e.g. proving that extension from some subspace to the whole space of some continuous map to $A$ isn't continuous). Also, $A$ is compact, which could also be something that we could use. So, assume that $A$ is a retract of $\mathbb R^2$. Then, $A$ is also an absolute extensor of normal spaces. $\mathrm{id}_A$ is a continuous map and therefore there exists a function $r\colon \mathbb R^2\rightarrow A$ such that $r|_A\equiv \mathrm{id}_A$. But I have not a clue what to do from here on.
At some point in yielding a contradiction, we should probably use that some chosen cover isn't locally finite, but I don't really know where that will help me.
The question was posted after an exam in my general topology class. I've now had my first geometric topology class and came back to answer this.
I have constructed a solution to this problem that is way more elementary than that of @freakish. Assume that $r\colon \mathbb R^2\rightarrow A$ is a retraction. Let $$ I_n=\{1\}\times [y_{m_n},y_{m_{n+1}}] \quad \forall n\in\mathbb N $$ where $y_m=\sqrt{m^2-(1-m)^2}$ and $m_n=1+\frac 1 n$. Let $n\in\mathbb N$. $I_n$ is a line that connects the endpoints of red arcs in the picture in the original post. Because $I_n$ is connected, $r(I_n)$ is too. Since $r(1,y_{m_n})=(1,y_{m_n})$ and also $r(1,y_{m_{n+1}})=(1,y_{m_{n+1}})$, there must hold $$ (A_n\cup A_{n+1})\cap([0,\infty)\times[0,\infty))\subset r(I_n), $$ thus $\exists a_n\in I_n$ such that $r(a_n)=(0,0)$. Now $$ \lim_{n\rightarrow\infty} a_n = (1,1) $$ but $$ (1,1)=r\left(\lim_{n\rightarrow\infty}a_n\right)=\lim_{n\rightarrow\infty}r(a_n)=(0,0). $$ Contradiction.