Where can I find a proof of the Swiss cheese theorem?

[Bestfrog]

Does anyone know where I can find a proof of this theorem?
Theorem: The Euclidean space $\mathbb{R}^2$ is not the union of nondegenerate disjoints circles.

 

[WWGD]

That is kind of old-school stuff, from the 50s-60s. This website : http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist&task=list is led (moderated) by someone specialized in that area. If not, try Mathstack exchange : https://math.stackexchange.com/questions , maybe use their search engine . Still, doesn't Baire category apply, if the union is countable?

 

[Infrared]

Consider a sequence of circles $S_i$ such that the disk bounded by $S_i$ contains $S_{i+1}$ and the radii $r_i$ of these circles tend to zero (this is possible because given a circle of radius $r$, the circle passing through its center has radius $\leq r/2$) and let $x_i$ be a point in the disk bounded by $S_i$. The sequence $x_i$ converges since $r_i\to 0$ and we find that the circle passing through the limit of this sequence must intersect all but finitely many of the $S_i$. Contradiction.

Edit: This seems too easy to be a proof of a substantial theorem, so I might be missing something.

Still, doesn't Baire category apply, if the union is countable?

If the union was countable, you could just use that the countable union of measure zero sets has measure zero.

 

[WWGD]

If the union was countable, you could just use that the countable union of measure zero sets has measure zero.

Seemed, by the type of problem, the OP wanted to address it from a purely topological perspective/approach.

 

[jostpuur]

I know a proof for a theorem that states that it is not possible to write the plane as a union of closed disks in such way that the interiors of the disks would be disjoint. In other words
$$\mathbb{R}^2 = \bigcup_{i\in\mathcal{I}} \bar{B}(x_i,r_i)$$
and
$$i,i'\in\mathcal{I},\; i\neq i'\quad\implies\quad B(x_i,r_i)\cap B(x_{i'},r_{i'})=\emptyset$$
lead to a contradiction, where $r_i>0$ for all $i\in\mathcal{I}$, the index set $\mathcal{I}$ can be arbitrary to start with,
$$B(x,r) = \big\{x'\in\mathbb{R}^2\;\big|\; \|x'-x\|<r\big\}$$
and
$$\bar{B}(x,r) = \big\{x'\in\mathbb{R}^2\;\big|\; \|x'-x\|\leq r\big\}.$$

 

[Svein]

I know a proof for a theorem that states that it is not possible to write the plane as a union of closed disks in such way that the interiors of the disks would be disjoint.

Often named a "Swiss cheese" (https://en.wikipedia.org/wiki/Swiss_cheese_(mathematics)).