Show that $\mathbb{R}^2$ can't be written as the union of disjoint topolocial circles

My attempt was to think about cohomology. But I think that there is some failure on my thought.

Suppose that the claim holds. Then $\mathbb{R}^2 = \cup_i S^1_i$. We know that $\mathbb{R}^2$ minus a circle has the comology isomorphic to $\mathbb{R}$. But does not make any difference take off a circle of such union, then the cohomology of $\mathbb{R}^2$ is not the trivial. What is a contradiction.

Solution 1:

Suppose you had such a decomposition. Pick a circle $C_1$. By Schoenflies it bounds a disc $D_1$. Inductively, pick a circle $C_n$ contained in $D_{n-1}$ such that the area (measure, if you like) of the disc $D_n$ it bounds is at most half the area of the area of $D_{n-1}$.

This is the key step in the proof, so we should carefully see why we can do this. For notational convenience I'm going to call $D=D_{n-1}$. First note that any disc properly contained in the interior of $D$ has strictly smaller area (its complement in the interior of $D$ is open hence has positive area). Now suppose I could not find a disc of arbitrarily small area bounded by one of our circles; then let the infimum of the areas of discs bounded by our circles be $t>0$. As above we see that we cannot actually achieve $t$, or else we would be able to achieve areas smaller than it by looking at circles inside the disc of area $t$. We will contradict this by showing that we can represent $t$.

Pick a sequence $S_n$ of circles with area at most $t+1/n$. Passing to a suvsequence if necessary and invoking the finite area of the disc I can assume the $S_n$ are nested downwards. Take the intersection of the discs they bound. By Cantor's intersection theorem this is nonempty; pick a point $x$ in it; because the circle containing $x$ must have been in the disc $S_n$ bounds for all $n$, that circle is contained in the infinite intersection; and so is the disc that circle bounds. But this disc must have area at most $t+1/n$ for all $n$, hence have area at most $t$, as desired.

The above discussion proved, recall, that we may always pick a circle $C_n$ contained in the previous, such that the area of the disc it bounds is at most half the previous. Now let's apply the previous argument again: take the intersection of all the $D_n$, it contains a disc, that disc has positive area, which is nonsense since its area is at most $\text{Area}(D_1)/2^n$ for all $n$. This contradicts our most crucial assumption: the existence of a decomposition into circles! Thus proven what we wanted to prove.

I see no way to do this using cohomological arguments. If you demand that the circles are a foliation of the plane it's much easier to prove impossibility.

Solution 2:

Suppose it can be written as a union of circles. Now consider a circle $C^1$. let $c_1$ denote its center. Then there exists a circle $C^2$ containing $c_1$. Let $c_2$ be the center of $C^2$. Then there exists $C^3$ a circle containing $c_2$. Now consider the sequence of center of circles {$c_i$}. Then since {$c_i$} is a bounded sequence , so by Bolzano-Weierstrass property of $\mathbb R^2$, it has a convergent sequene which will converge to $c$. Now there should exists a cicle $C$ contatining $c$. But then $C$ will intersect $C^n$ for some large $n$( Because observe that radius of $C^{i+1}\leq 1/2$ radius of $C^i$). Contradiction.