I'm assuming has the usual topology. So, we know that , right? But, since is open there exists some neighborhood of such that . But, whatever neighborhood is it hits points strictly less than and so points such that . So, let be one of those points. Then, the last part gotten since every quotient map is surjective.

You can do this.(b) Deduce that the open sets in R/R are ∅, R/R and the image sets p(−∞, n] , where n ∈ Z.

Of course the map isn't continuous. For, given any we have that is open and so with the subspace topology is discrete. And so, if were continuous then should be open, but(c) Consider the map I : R → Z, x ↦→ I(x) . Is the map I continuous (when Z is endowed with the induced topology) ?

Prove that I deﬁnes a bijection I : R/R → Z. What is the topology on Z making I a homeomorphism ?

I think you have your map a little screwed up. So you're saying that (x)\mapsto I(x)" alt="I:\mathbb{R}/\sim\to\mathbb{Z}(x)\mapsto I(x)" />, right? It doesn't go without saying that the map is well defined. But, if then for some and from where it follows that . Now, to see it's an injection suppose that then and just as before and so . To see it's a surjection you merely must note that .

I would say you need to give it the quotient topology determined by

So, to see now that is continuous we need merely note that a mapping out of a quotient space is continuous if and only if it's composition with the definining quotient map is continuous. But, and of course this is continuous, it is the quotient map used to define the topology on !

So, why is it's inverse continuous? Well, let's compose it with the quotient map defined in the making of . We get....I'll let you finish this.