# $\bigvee _{n\in\mathbb{N}}\mathbb{S}^1$ and its homeomorphisms

How would you prove there doesn't exist any $X\subset\mathbb{R}^2 : X \cong \bigvee _{n\in\mathbb{N}}\mathbb{S}^1$?

By now I've realized that since $X\subset\mathbb{R}^2$, $X$ satisfies the first countability axiom, and if there existed an homeomorphism between $X$ and $\bigvee _{n\in\mathbb{N}}\mathbb{S}^1$, $\bigvee _{n\in\mathbb{N}}\mathbb{S}^1$ would have to satisfy the 1CA as well. That's where I get stuck, I don't know how to prove that set doesn't satisfy it.

NOTATION: $\bigvee _{n\in\mathbb{N}}\mathbb{S}^1:={(\coprod_{n\in\mathbb{N}}\mathbb{S}^1})/{\sim}$, where $\sim$ identifies each sphere's base point (in this case $(1,0)$ ) into a single one.

• Do you mean ‘first countability’? May 20, 2015 at 16:42
• Yes, first countability sorry
– GSF
May 20, 2015 at 16:49
• Are you sure this is true? What if $X$ is the union of $S_n$ where $S_n$ is the circle given by the equation $(x-n)^2 + y^2 = n^2$? (Think of starting with a small circle, and drawing larger and larger circles around it, all meeting at some predetermined point.) May 20, 2015 at 17:09
• @JasonDeVito I guess you could find a continuous and bijective mapping between $X$ and the union of those sets, but what about the inverse of that mapping, would it be continuous? I think that would only be true for finite values of $n$.
– GSF
May 20, 2015 at 17:19
• May 20, 2015 at 17:26

suppose there exists $X\subset\mathbb{R}^2 : X \cong \bigvee _{n\in\mathbb{N}}\mathbb{S}^1$...let $f$ be such a homeomorphism ...then there is a point $a \in \mathbb{R^2}$ s.t base point will map to $a$...and $f$ maps each circle to some curves on $\mathbb{R^2}$...now observe that $f$ is a open map too...if I consider $1/n$ length of open segment passing through $a$ of the curve corresponds to the $nth$ cirlce (if possible) as $l_n$ (i.e $|l_n|=1/n$ and $a\in l_n$)..then $\cup_{n\geq 1} f^{-1}(l_n)$ is open in $\bigvee _{n\in\mathbb{N}}\mathbb{S}^1$ (simply follows from the definition of cw-complex)... but $f(\cup_{n\geq 1} f^{-1}(l_n))= \cup_{n\geq1}l_n$ is not open in $X$...because you cannot find any open set inside this set.
• Do you mean that $\bigcup_{n ≥ 1} l_n$ is not open in $X$? May 20, 2015 at 18:45