4
$\begingroup$

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.

$\endgroup$
9
  • $\begingroup$ Do you mean ‘first countability’? $\endgroup$ May 20, 2015 at 16:42
  • $\begingroup$ Yes, first countability sorry $\endgroup$
    – GSF
    May 20, 2015 at 16:49
  • 1
    $\begingroup$ 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.) $\endgroup$ May 20, 2015 at 17:09
  • 2
    $\begingroup$ @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 $n$. $\endgroup$
    – GSF
    May 20, 2015 at 17:19
  • 2
    $\begingroup$ Related: math.stackexchange.com/questions/111489/… $\endgroup$
    – Dejan Govc
    May 20, 2015 at 17:26

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Do you mean that $\bigcup_{n ≥ 1} l_n$ is not open in $X$? $\endgroup$
    – user87690
    May 20, 2015 at 18:45
  • $\begingroup$ yes...sorry you are correct $\endgroup$ May 20, 2015 at 20:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.