# Is N the cts image of the Sorgenfrey line?

I have this question:

Prove/disprove: The set of natural numbers (including zero) with usual topology is the continuous image of the Sorgenfrey line.

Can't we take the map $g: \mathbb{R}_{l} \rightarrow \mathbb{N}$ given by $g(x) = |[x]|$ ? (i.e the absolute value of the floor function).

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The OP seems to have a pretty good handle on the question and maybe just needs to think it through to be fully satisfied.

But here is a slightly different approach which makes it even more clear (to me) that the OP's method will work. This was borne out of the remark I made (to myself) upon reading the OP's clarification that the natural numbers include zero (you're damn right they do, by the way): namely, it certainly doesn't matter, because any two countably infinite discrete spaces are homeomorphic.

Taking that one step further, it is clearly enough to realize $\mathbb{Z}$ with the discrete topology as a continuous image of the Sorgenfrey line: having done this, compose with any homeomorphism (i.e., bijection!) from $\mathbb{Z}$ to $\mathbb{N}$. (Or, in fact, with any surjection, as the OP has done.) For this, literally take the greatest integer function. The preimage on any given basis element -- i.e., a singleton set $\{n\}$ -- is the half-open interval $[n,n+1)$. I don't keep too much information about the Sorgenfrey line in my head, but I'm pretty sure those sets are open!

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Yes we can. Is your question how to show that it is continuous? Is the inverse image of $\{k\}$ open in $\mathbb{R}_l$?

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yep, sorry, we only need to check what happens to singletons right? (because N is discrete). – user6253 Jan 26 '11 at 1:57
@Liz: Yes. In general is it enough to check on a basis. – Jonas Meyer Jan 26 '11 at 2:02