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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.

Denis
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3 Answers3

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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 $x$.

copper.hat
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  • Thanks for your answer! What do you mean by "connected domain"? Maybe I was taught a different term for it? – Denis Nov 12 '12 at 23:40
  • This is what I mean http://en.wikipedia.org/wiki/Connected_space. I added an alternative approach using the intermediate value theorem. – copper.hat Nov 12 '12 at 23:42
  • Thank you! :) Much appreciated! – Denis Nov 12 '12 at 23:45
  • @copper.hat is it possible to use the properties of the connected domain like this. We pick $a \in X$. Let $G_1$ be the set of all points in $X$ for which $f(x)=f(a)$. Now due to continuity on $X$, some small $|x-x_0|<\delta$ in $G_1$ exists, the same holds for $G_2=X \setminus G_1$. So both $G_1$ and $G_2$ are open. Moreover, $f(x_0)=f(a)=f(x)$ for $|x-x_0|<\delta$. Since $a \in X$ is non empty, then $G_2=\emptyset$. On the set $G_1$. So $f(x)=f(a)$ on $X$ – Alexander Cska Apr 20 '22 at 18:13
  • @AlexanderCska Yes, that is a good approach. – copper.hat Apr 20 '22 at 20:53
  • @copper.hat I came about this theorem when studying the complex log. There it is used to prove that regular branches of the log differ by $2ki\pi $. $e^{F_1(z)}=e^{F_2(z)}=f(z)$ then $F_1(z)-F_2(z)=k(z)2i\pi $. Or $k(z)=k$. How does the proof applies to this concrete function? I have hard time visualising the proof. – Alexander Cska Feb 03 '24 at 19:57
  • @AlexanderCska The first proof applies. Presumably the domain of $F_1,F_2$ is connected, hence $k$ maps a connected set into a connected set and the only connected subsets of the integers are the empty set and singletons. – copper.hat Feb 04 '24 at 04:22
  • @copper.hat in the book they use a lemma that if for the values of a continious $k(z)$ on some domain $D$ function, $|w_1 - w_2|>=d>0$ holds, then $k(z)$ is constant. Of course $w_1 \neq w_2$ and they use the properties of simply connected. I don't understand how $k(z)=k$ is a constant? $k$ can be any integer. – Alexander Cska Feb 04 '24 at 12:09
  • @AlexanderCska you can adapt the third approach above, use the fact that $D$ is path connected and let $f(z)=|k(z)-w_1|$. If you need further elaboration, consider asking another question. – copper.hat Feb 04 '24 at 21:16
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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.

Brian M. Scott
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Hint

$$f^{-1}\big(\{f(0)\}\big)$$ is clopen in $\mathbb R$.

Surb
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