Prove that any continuous integer-valued function of a real variable is constant. I'm stuck on a question which asks to prove that any continuous integer-valued function of a real variable is constant. In my lecture notes we are told that a map is continuous if the preimage of any open set is open.
How do I use this to answer the question?
A full answer would be greatly appreciated as I can't seem to figure out how to solve it in my 
notes, in analysis books or anywhere online.
 A: I’m assuming that function of a real variable means a function defined on an interval, a ray, or the whole real line, as the result isn’t true otherwise. If you’re allowed to use familiar results from calculus, you can simply appeal to the intermediate value theorem: if $a<b$, $f(a)=m$, and $f(b)=n\ne m$, pick any non-integer $c$ between $m$ and $n$ and note that if $f$ is continuous, there must be some $x\in(a,b)$ such that $f(x)=c\notin\Bbb Z$.
If you’re not allowed to use that, the proof is substantially harder, since in effect you have to prove a special case of that theorem.
A: It is not necessarily true unless the domain is connected.
If the domain $X$ is connected and $f$ is continuous, then $f(X)$ is also connected. The only connected subsets of the integers are sets containing at most one point. Hence $f(X)$ (which is presumably non-empty) has exactly one point and $f$ is constant on $X$.
Another approach is to suppose that $y_1=f(x_1), y_2=f(x_2)$. Suppose $y_1\leq y_2$ without loss of generality. Then by the intermediate value theorem, $f$ must take all the values in $[y_1,y_2]$. However, since the range is the integers, this can only be true if $f$ only takes one value.
Here's a third approach: Since $f$ is continuous, for each $x_0$ we can find a $\delta>0$ such that if $|x-x_0|<\delta$, then $|f(x)-f(x_0)| < \frac{1}{2}$. Since $f$ is integer valued, this means that $f(x) = f(x_0)$ for all $|x-x_0|<\delta$.
Now let $t_+ = \sup \{x | f(x) = f(0)\}$. If $t_+<\infty$, then by continuity, we must have $f(t_+) = f(0)$, and by the previous paragraph, there is a $\delta>0$ such that $f(x) = f(t_+)$ for all $x$ satisfying $|x-t_+| < \delta$, which contradicts the definition of $t_+$. Hence $t_+ = \infty$. Applying the same approach to $t_- = \inf \{x | f(x) = f(0)\}$ shows that $t_- = -\infty$, hence $f(x) =f(0)$ for all $x$.
A: Hint
$$f^{-1}\big(\{f(0)\}\big)$$ is clopen in $\mathbb R$.
