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I know that every function $f: \mathbb{Z} \to \mathbb{R}$ is continuous.

This can be proven by Weierstrass' $\varepsilon$-$\delta$ criterion. If $f:E\subseteq \mathbb{R} \to \mathbb{R}$ is continuous at $x_0 \in E$, then $$\forall \varepsilon > 0 \exists \delta > 0 \forall x \in E \left( \left| x - x_0\right| < \delta \to \left| f(x) - f(x_0)\right| < \varepsilon \right)$$ Let $E := \mathbb{Z}$ and $\varepsilon > 0$, $x_0 \in E$ be fixed. Taking $\delta := 1$, we get, that the only point which fulfills $\left| x - x_0\right| < \delta$ is $x_0$ itself, but then trivially $$\left| f(x_0) - f(x_0)\right| = 0 < \varepsilon$$ holds. Since $x_0 \in E$ was arbitrary, $f$ is continuous on $E$.

But how can one prove/disprove that there exists only one continuous function:

$$f:\mathbb{R} \to \mathbb{Z}$$

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    $\begingroup$ Back up, before you attempt a proof, why do you think this is true? $\endgroup$ – Chris Culter Jan 7 '20 at 17:41
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    $\begingroup$ There are in fact infinitely many continuous functions $\;\Bbb R\to\Bbb Z\;$ (with the usual topology and the inherited one, respectively), and each one of them pretty boring because $\;\Bbb R\;$ is connected... $\endgroup$ – DonAntonio Jan 7 '20 at 17:44
  • $\begingroup$ @DonAntonio Thanks for your reply. My assumption was wrong because I couldn't find a counterexample. But I don't understand what you mean with $\mathbb{R}$ is connected. Can you elaborate? $\endgroup$ – JavaTeachMe2018 Jan 7 '20 at 17:45
  • $\begingroup$ @Java In this case it is the same as saying that $\;\Bbb R\;$ is path-connected...Anyway, being connected means that we cannot write $\;\Bbb R=A\cup B\;$ , with $\;A\cap B=\emptyset\;$ and $\;A,B\;$ both non empty and open. $\endgroup$ – DonAntonio Jan 7 '20 at 17:49
  • $\begingroup$ The point to connectedness here is that if $X$ is connected and $f$ is continuous then $f(X)$ is connected. So here $f(\Bbb R)$ is a connected subset of $\Bbb Z$, hence it contains only one point, which is to say $f$ is constant. $\endgroup$ – David C. Ullrich Jan 7 '20 at 18:04
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Hint: Is it not the case that $f:\mathbb{R}\to \mathbb{Z}$ given by $f(x)=1$ and $g:\mathbb{R}\to \mathbb{Z}$ given by $g(x)=0$ are both continuous?

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Any constant function $\mathbb{R} \to \mathbb{Z}$ is continuous, so this claim is definitely false.


But these are the only counterexamples:

Indeed, suppose we are given a continuous function $f: \mathbb{R} \to \mathbb{Z}$. Since $\mathbb{R}$ is connected, we have that $f(\mathbb{R})$ is a connected subspace of $\mathbb{Z}$. The only connected subspaces of $\mathbb{Z}$ are singeltons, so $f$ must necessarily be constant.

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There are many ways to show that a continuous $f:\Bbb R\to \Bbb Z$ must be constant. We can consider $f$ to be continuous from $\Bbb R$ into $\Bbb R$ and use some general properties of all continuous real functions. For example, the Intermediate Value Property: Suppose $f:\Bbb R\to \Bbb R$ is continuous and not constant. Then there are $x,x'$ with $f(x)\ne f(x').$ Now for every $y$ between $f(x)$ and $f(x'),$ there exists $x''$ between $x$ and $x'$ such that $f(x'')=y.$ But there is always some $y$ between $f(x)$ and $f(x')$ with $y\not \in \Bbb Z;$ therefore it is not possible that $\{f(x''):x''\in \Bbb R\} \subseteq \Bbb Z.$

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