# The closure of the union of growing circles in $\Bbb R^2$

My question is related to this question: About two space with infinite circles.. But what I am curious about is not actually the same with the linked one, so I make here a new question.

For each positive integer $$n$$, let $$C_n$$ denote the circle in $$\Bbb R^2$$ of radius $$n$$ centered at $$(n,0)$$. Then the $$C_n$$'s meet each other at the origin $$(0,0)$$. Let $$X$$ be their union $$\bigcup_{n=1}^\infty C_n$$. Then evidently the closure of $$X$$ equals $$X$$ union the $$y$$-axis. What I want to know is whether $$X$$ and $$\bar{X}$$ are homotopy equivalent or not.

What I know is: $$X$$ is homotopy equivalent, but not homeomorphic, to the wedge sum of countably inifnitely many copies of $$S^1$$. Also $$X$$ is certainly not a deformation retract of $$\bar{X}$$.

Let us regard $$X$$ as subspace of $$\mathbb C$$.

That $$X$$ and $$\overline X$$ are not homeomorphic is easy to see: The space $$\overline X \setminus \{i\}$$ has two path components (proof left as an exercise!), but $$X \setminus \{0\}$$ has infinitely many path components and for $$z \ne 0$$ the space $$X \setminus \{z\}$$ has one path component.

We shall show that the inclusion $$i : X \to \overline X$$ is a homotopy equivalence.

The idea for constructing a homotopy inverse is this: Map the vertical line $$L = \{z \in \mathbb C \mid \text{Re}(z) = 0 \}$$ and the "left half circles" $$C_n^l = \{ z \mid \lvert z-n \rvert = n, \text{Re}(z) \le n \}$$ to $$0$$ and map the "right half circles" $$C_n^r = \{ z \mid \lvert z-n \rvert = n, \text{Re}(z) \ge n \}$$ onto the full circles $$C_n$$ such that the two boundary points $$n(1 \pm i)$$ of $$C^r_n$$ are sent to $$0$$. Let us make this precise.

Define $$M = L \cup \bigcup_{n=1}^\infty C_n^l$$ and $$H : M \times I \to \overline X, H(z,t) = \begin{cases} n \left(1 + \dfrac{tz-n}{\lvert tz-n \rvert} \right) & z \in C^l_n \\ tz & z \in L \end{cases}$$ The geometric idea for this definition is this: Given $$z \in C^l_n$$, consider the line $$L(z,t)$$ through $$n$$ and $$tz$$ (which lies on the secant segment connecting $$0$$ and $$z$$). It intersects $$C^l_n$$ in a single point $$H(z,t)$$. Note that by construction $$H( C_n^l \times I) \subset C_n^l$$ and $$H(L \times I) \subset L$$. Moreover $$H(z,0) = 0$$ and $$H(\overline z,t) = \overline {H(z,t)}$$. We shall show later that $$H$$ is continous.

The map $$f_n : [0,\pi] \to C_n, f_n(s) = n(1+e^{is})$$, is an embedding whose image is the closed upper half circle. Let $$R = \bigcup_{n=1}^\infty C_n^r$$. The $$C^r_n$$ are open subspaces of $$R$$. Therefore $$G : R \times I \to \overline X, G(z,t) = \begin{cases} G_+(z,t) = f_n\left(\frac{2f_n^{-1}(z)f_n^{-1}(H(n(1+i),t)}{\pi}\right)) & z \in C^r_n , \text{Im}(z) \ge 0 \\ \overline{G_+(\overline z,t)} & z \in C^r_n , \text{Im}(z) \le 0 \end{cases}$$ is continuous. The geometric idea for this definition is this: Pull the two the boundary points $$n(1 \pm i)$$ of $$C^r_n$$ to the left along the circle $$C_n$$ until they reach the points $$H(n(1 \pm i),t)$$. This induces a deformation of $$C^r_n$$ which takes place inside $$C_n$$.

Since $$H, G$$ are defined on closed subspaces of $$\overline X \times I$$ whose union is $$\overline X \times I$$ and $$H, G$$ agree on the intersection of these sets, we get a continuous homotopy $$K : \overline X \times I \to \overline X .$$ Note that $$K(X \times I) \subset X$$ and $$K(z,1) \in X$$ for all $$z \in \overline X$$. Let $$\rho : \overline X \to X, \rho(z) = K(z,1).$$ Clearly $$i \circ \rho$$ is homotopic via $$K$$ to $$id$$, similarly $$\rho \circ i$$ is homotopic via $$K \mid_{X \times I}$$ to $$id$$.

This proves that $$i$$ is a homotopy equivalence.

Let is finally verify that $$H$$ is continuous. This is technically somewhat nasty.

Clearly $$H$$ is continuous in all points of $$(M \setminus L) \times I$$ because $$M \setminus L = \bigcup_{n=1}^\infty \left(C_n^l \setminus \{0\}\right)$$ and the sets $$C_n^l \setminus \{0\}$$ are open in $$M$$. Let us show that $$H$$ is continuous in all points $$(\zeta,\tau) \in L \times I$$. Since $$H \mid_{L \times I}$$ is continuous, it suffices to consider a point $$(\zeta,\tau) \in L \times I$$ and to prove that for each $$\epsilon > 0$$ one has $$\lvert H(z,t) - \tau \zeta \rvert < \epsilon$$ for all $$(z,t) \in M \setminus L$$ which are sufficiently close to $$(\zeta,\tau)$$. To do so, it clearly suffices to show that $$\lvert H(z,t) - t z \rvert < \epsilon \text{ if } (z,t) \text{ is sufficiently close to } (\zeta,\tau) .$$ Let $$n$$ denote the unique index such that $$z \in C^l_n$$ and write $$z = x +iy$$. The point $$H(z,t)$$ was obtained as the intersection of the line $$L(z,t)$$ with $$C^l_n$$. For $$z \ne n(1 \pm i)$$ the line $$L(z,t)$$ also intersects $$L$$ and it is easy to see that the intersection point is $$g(z,t) = i\dfrac{nty}{n-tx}$$. Note that for $$z \in C^l_n \setminus \{n(1+i), n(1-i)\}$$ we have $$tx \le x < n$$. Clearly $$\lvert H(z,t) - t z \rvert \le \lvert g(z,t) - t z \rvert$$. Straightforward calculations show that $$\lvert g(z,t) - t z \rvert^2 = t^2x^2 + \dfrac{t^4x^2y^2}{(n - tx)^2} \le x^2(1 + \dfrac{y^2}{(n - tx)^2}).$$ Write $$\zeta = i\eta$$ with $$\eta \in \mathbb R$$. Then for $$\lvert z - \zeta \rvert < \frac{1}{2}$$ we have $$0 \le tx \le x \le \sqrt{x^2 +(y - \eta)^2} = \lvert z - \zeta \rvert < \frac{1}{2} < n$$, thus we are in the situation $$z \ne n(1 \pm i)$$. Moreover $$n - tx \ge n - x \ge 1 - \frac{1}{2} = \frac{1}{2}$$ and $$\lvert y - \eta \rvert \le \lvert z - \zeta \rvert < \frac{1}{2}$$. In particular $$\lvert y \rvert < \lvert \eta \rvert + \frac{1}{2}$$ and therefore $$\lvert g(z,t) - t z \rvert^2 \le x^2(1 + 4(\lvert \eta \rvert + \frac{1}{2})^2)$$ which completes the proof because $$x \to 0$$ as $$z \to \zeta$$.

• This is a very nice answer. +1 – Tyrone Oct 27 '20 at 18:58

## Never mind, this does not work.

$$X$$ and $$\overline{X}$$ are homotopy equivalent by the following:

Let $$C_0$$ be the circle of radius $$1$$ and in general let $$C_n$$ denote the circle of radius $$n+1$$. Consider the homotopy that, on the interval $$[0,\tfrac{1}{2}]$$, shrinks $$C_0$$ so as to have radius $$(1-2t)+(2t)tan^{-1}(1)$$ and leaves the remaining $$C_n$$ and the line $$\mathbb{R}\times \{0\}$$ fixed. More generally, on the interval $$[\tfrac{2^{n}-1}{2^n},\tfrac{2^{n+1}-1}{2^{n+1}}]$$ the map deforms $$C_n$$ into the circle of radius $$\Big(1-2^{n+1}\Big(t-\frac{2^n-1}{2^n}\Big)\Big)(n+1)+\Big(2^{n+1}\Big(t-\frac{2^n-1}{2^n}\Big)\Big)tan^{-1}(n+1)$$ leaving everything else fixed. At the end of the homotopy, all of our circles will be bunched up inside of a missing circle having radius $$\frac{\pi}{2}$$. Since $$\mathbb{R}\times \{0\}$$ is fixed at each stage, the restriction of the homotopy to this subset is the identity. With the circles unable to interfere, we can crush $$\mathbb{R}\times \{0\}$$ and then launch all the circles back to their original positions with a homotopy constructed from the tangent function.

• This does not work. Shrinking the $C_n$ is possible in $\mathbb R^2$, but you need a homotopy taking place in $\overline X$. It is impossible to find a homotopy shifting all points of $C_n$ (except $0$) into $\overline X \setminus (C_n \setminus \{0\})$. – Paul Frost Oct 25 '20 at 22:51
• Oh my, this really was wrongheaded. Thank you for pointing out my error. – Gary D Oct 25 '20 at 23:25