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

Question

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.

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.) — Jason DeVito - on hiatus, May 20 '15 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 '15 at 17:19
Related: http://math.stackexchange.com/questions/111489/the-union-of-growing-circles-is-not-homeomorphic-to-wedge-sum-of-circles?rq=1 — Dejan Govc, May 20 '15 at 17:26
You are right, the $\bigvee_{n ∈ \mathbb{N}} \mathbb{S^1}$ is not first countable. It is because you can define a neighborhood of the basepoint independently on each circle. — user87690, May 20 '15 at 18:47

Anubhav Mukherjee · Answer 1 · 2015-05-20T20:01:05.997

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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.

edited May 20 '15 at 20:01

answered May 20 '15 at 17:31

Anubhav Mukherjee

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Do you mean that $\bigcup_{n ≥ 1} l_n$ is not open in $X$? – user87690 May 20 '15 at 18:45
yes...sorry you are correct – Anubhav Mukherjee May 20 '15 at 20:00

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

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