Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $f:\mathbb R\times \mathbb R\to \mathbb R/\mathbb Z\times \mathbb R/\mathbb Z$ where the domain is given the usual topology and the latter the quotient topology. Why then is the restriction $f|_S:S\to f(S)$, where $S=\{(x,\sqrt{5} x): x\in \mathbb R\}$ not continuous with continuous inverse?

If I am not wrong, the quotient topology is comprised of sets whose preimages are open.

I can prove that $f|_S^{-1}$ is well-defined by showing that $f|_S$ is bijective, but I don't know why they are not continuous.

share|improve this question
    
are you asking for a proof or for intuition? –  Tim kinsella May 19 '12 at 10:50
    
the intuition is that $f|_S$ maps distant points in $S$ to nearby points in the codomain so that the $f|_S^{-1}$ would have to tear those nearby points away from each other. –  Tim kinsella May 19 '12 at 10:53

1 Answer 1

up vote 0 down vote accepted

Let $\alpha=\frac12\sqrt2$, and let $p_n=\left\langle n\alpha,n\alpha\sqrt5\right\rangle\in S$ for $n\in\Bbb N$; then $\{f(p_n):n\in\Bbb N\}$ is dense in $\left(\Bbb R/\Bbb Z\right)^2$. Thus, there is a strictly increasing sequence $\langle n(k):k\in\Bbb N\rangle$ in $\Bbb N$ such that $$\left\langle f(p_{n(k)}):k\in\Bbb N\right\rangle\to p_0$$ in $\left(\Bbb R/\Bbb Z\right)^2$. Let $g=f\upharpoonright S$, and let $h=g^{-1}$. If $h$ were continuous, we’d have $$\left\langle p_{n(k)}:k\in\Bbb N\right\rangle=\left\langle h\left(f(p_{n(k)})\right):k\in\Bbb N\right\rangle\to h(p_0)=\langle 0,0\rangle\;,$$ which is absurd.

The density of $\{f(p_n):n\in\Bbb N\}$ in $\left(\Bbb R/\Bbb Z\right)^2$ follows from the fact that $1,\alpha$, and $\alpha\sqrt5$ are pairwise incommensurate: none of them is a rational multiple of another.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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