Prove that there exists continuous function of Sorgenfrey line to space $\mathbb{N}$... Prove that there exists continuous function of Sorgenfrey line to space $\mathbb{N}$ with induced topology from euclidan topology on $\mathbb{R}$.
I don't know if I understand the continuity definition properly.
I have two topological spaces: $(\mathbb{R}, \tau_1) $ and $(\mathbb{N}, \tau_2)$, where $\tau_1$ is Sorgenfrey line and $\tau_2$ is my induced topology.
I want to define a function: $f: \mathbb{R} \rightarrow \mathbb{N}$
Let $U \in \tau_1$. We know that $U$ is an interval such that $U=[a,b)$ for $a<b$.
Let $f(U)=\{n\in \mathbb{N}: a\leq n < b \} $
Of course our product is in $\tau_2$.
No I will be less formal, I just want to know if my thinking is correct.
Let's take an example: $W=\{ 2,3,4,5\}\in\tau_2$. Then $f^{-1}(W)=\{[2,5+i): i\in(0,1]\}\in\tau_1$  ?
I don't know how to prove it yet, but it is pretty obvious $\forall_{V\in\tau_2} f^{-1}(V)\in\tau_1$
Is my definition of $f$ correct? Do I understand the definition properly?
 A: It appears that you may be thinking in the right general direction, but what you’ve written has some major problems.

Let $U\in\tau_1$. We know that $U$ is an interval such that $U=[a,b)$ for $a<b$.

This is not true: what we know is that either $U=\varnothing$, or $U$ is a union of intervals of the form $[a,b)$.

Let $f(U)=\{n\in\Bbb N:a\le n<b\}$.

The function $f$ is supposed to be a function from $\Bbb R$ to $\Bbb N$, not from $\tau_1$ to $\tau_2$. You need to tell us what $f(x)$ is for arbitrary $x\in\Bbb R$.
Can you convert your idea into an actual function $f:\Bbb R\to\Bbb N$? I’ve left one possible answer in the spoiler-protected block below.

 $f(x)=\lfloor x\rfloor$, the greatest integer in $x$.

A: The induced topology of $\mathbb{N}$ is the discrete topology. So in order for a function $f: (\mathbb{R}, \tau_1) \rightarrow \mathbb{N}$ to be continuous it is necessary and sufficient that for all $n$, $f^{-1}[\{n\}]$ be open in the Sorgenfrey topology. 
Consider the function $f$ that sends all $x < 0$ to $0$ and every other $x$ to $\max \{n: n \le x \}$ (the floor of $x$). So $f[[0,1)] = \{1\}$, e.g. Show that this works. 
