Any continuous map $f:\mathbb{R}\to\mathbb{Z}$ is constant

Question

Not really sure how to proceed with this one. Can anyone help?

The question is:

Show that any continuous map $f: \mathbb{R}\to\mathbb{Z}$ is constant.

Answer 1

Suppose the continuous map $f: X\to \mathbb{Z}$ is not constant. Then there exist $a,b\in X$ such that $f(a)<f(b)$. Let $A=\{n\in\mathbb{Z}\mid n\leq f(a)\}$ and $B=\{n\in\mathbb{Z}\mid n > f(a)\}$. Both $A$ and $B$ are open sets, $A\cup B=\mathbb{Z}$ and $A\cap B=\emptyset$. Therefore the preimage of each is open, $f^{-1}(A)\cup f^{-1}(B)=X$ and $f^{-1}(A)\cap f^{-1}(B)=\emptyset$. Therefore the domain $X$ cannot be a connected space.

Answer 2

$f$ is continuous, thus by the intermediate value theorem, $f(\mathbb{R})$ is an interval. If $f$ was not constant, $f(\mathbb{R})$ would contain at least two different values, let $x< y$ in $f(\mathbb{R})$, since $f(\mathbb{R})$ is an interval, $]x,y[\subset f(\mathbb{R})\subset\mathbb{Z}$ which is not.

Answer 3

Use the basic definition : Say $f(c)=n$ Now choose $\epsilon <1$, so there is a $\delta$ such that $|x-c|<\delta\implies |f(x)-f(c)|<\epsilon <1$ which contradicts the fact that $|f(x)-f(c)|\geq 1$, since $f(c), f(x)$ are both integers, so unless they are equal, the absolute difference must not be smaller than $1$.

Answer 4

Let $x$ be a real number. Note that $\{f(x)\}$ is a closed-open set in $\mathbb{Z}$. So $f^{-1}(\{f(x)\})$ must be a closed-open set of $\mathbb{R}$. Because it's not empty (it contains $x$) and $\mathbb{R}$ is connected, it follows that $f^{-1}(\{f(x)\})=\mathbb{R}$. $f$ is constant.

This only uses basic facts about connexity (the definition and $\mathbb{R}$ being connected) but if you want to go faster, you can just say that if $f$ is not constant, then $f(\mathbb{R})$ is a set containing at least two elements. It's not connected but continuous maps maintain connexity and $\mathbb{R}$ is. There's a contradiction.