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I'm trying to solve two topology assignment problems as in the images and I'm a bit lost how to solve them:enter image description here

For Problem 1: I did a bit of search on the internet, but found that it's isomorphic to $\mathbb{R}$ as a group, but I found that it's not homeomorphic to $\mathbb{R}?$ So I'm having difficulty identifying $\mathbb{R}/\mathbb{Q}$ to any familiar space, as the assignment asks?

For problem 2b): I think it's topologically the half-open interval $[0, \infty),$ as the equivalence relation identifies all the points lying in any circle with center at the origin, i.e. $\forall r \ge 0,$ all $(x,y)$ so that $x^2+y^2=r^2$ are identified. So the quotient space should consist of all the classes $[(x,y)]$ with different values of $x^2 + y^2.$ These different values are just $[0, \infty),$ hence my answer. Is it correct?

For problem 2a): I feel it'd be just $\mathbb{R},$ by just using the above argument and noting that $y-x^3$ can take any real values. Is this correct?

Stat_math
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    Does this help? https://math.stackexchange.com/questions/24098/quotient-space-mathbbr-mathbbq – balddraz Nov 17 '20 at 09:57
  • @0XLR Thanks for the link - I saw the hint there given by Ryan Budney and now it seems to me that the quotient space $R/Q$ is just the irrationals $R-Q$ union with the single set ${0}$. I'm using the logic here that any two rationals differ by a rational, so all of them are in the same coset, calling it the coset $[0].$ Next, while there're irrationals differing by rationals, there certainly is $(R\Q)$-many irrationals differing by irrationals. So it seems there's a bijection from the quotient space $R/Q$ onto the irrationals union 0. I'm still not sure why it's a homeomorphism though. – Stat_math Nov 17 '20 at 10:11
  • @0XLR I think in his bhint, Ryan Budney refers to a discrete topology, but I'm trying to see why $R/Q$ got the discrete topology? – Stat_math Nov 17 '20 at 10:13

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For the first problem consider any open interval $(a,b)$ in $\Bbb R$: it intersects every $\sim$-equivalence class (why?), so every non-empty open set in $\Bbb R$ intersects every $\sim$-equivalence class. Now let $q:\Bbb R\to\Bbb R/\!\sim$ be the quotient map, and suppose that $U$ is a non-empty open set in the quotient space. By definition $q^{-1}[U]$ is open in $\Bbb R$, and clearly it must be non-empty. It also must contain every $\sim$-class that it intersects, so what set must it be?

Your answers to the second problem are correct. The equivalence classes in (a) are the graphs of $y=x^3+a$ for $a\in\Bbb R$, and a set of these graphs is open in the quotient iff the corresponding set of constants $a$ is open in $\Bbb R$. And as you saw, the situation in (b) is similar.

Brian M. Scott
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