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(a) Prove if $f: M \to \mathbb{R}$ is a continuous function such that all values are integers, then $f$ is constant whenever $M$ is connected

(b) What if all values are irrationals?

My attempt.

For (a), since $f$ is continuous, $f(M) \subset \mathbb{Z}$ is connected. But every metric space countable is disconnected and so, $f(M)$ is disconnected or constant. Also, $f(M)$ is connected because $f$ is continuous, then $f$ is constant.

Is correct?

For (b), I dont know. Equivalently to item (a), every connected metric space with at least $2$ points is uncountable, so I think maybe $f$ not need to be constant. I appreciate any help!

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    $\begingroup$ Irrationals, as subspace of the real line, are totally disconnected. If $f$ si continuous and $M$ connected, $f(M)$ is also connected. Hence $f$ must be constant. $\endgroup$ – Dog_69 Dec 22 '18 at 19:09
  • $\begingroup$ I got it! Thank you! $\endgroup$ – Corrêa Dec 22 '18 at 19:38
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    $\begingroup$ you can also think of it like that. if $f$ takes at least two different values $i_1$ and $i_2$ irrationals, then since there exists a rational $q$ in $]i_1,i_2[$ and $f$ continuous then $q$ should be reached by $f$ which is a contradiction. $\endgroup$ – zwim Dec 22 '18 at 20:35
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    $\begingroup$ @Dog_69 Why not an official answer? $\endgroup$ – Paul Frost Dec 22 '18 at 22:50
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    $\begingroup$ @Dog_69 I think you should do so. Answering a question in form of a comment does not make it visible in the queston queue that it has alredy been answered. If you click at "Unanswered" in the upper left corne of this page, you will read something like "235,204 questions with no upvoted or accepted answers ", the number of course growing. If you look at these questions, you will see that a great many are actually answered in comments. You also have the option to make your answer a "community wiki" (see math.stackexchange.com/help/privileges/edit-community-wiki). $\endgroup$ – Paul Frost Dec 24 '18 at 8:09
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Following the helpful comment of @PaulFrost I have decided to post my comment as an answer.

So, the argument runs as follows:

The irrationals, with the subspace topology inherited from the real line, are totally disconnected. To see that, let me denote $I=\mathbb R\setminus \mathbb Q$ and let $A\subseteq I$ be a subset with at least two different points; given $A,b\in A$, $a\neq b$, there exists $q\in\mathbb Q$ such that $a<q<b$ and thus the sets $(-\infty,q)\cap A$ and $(q,+\infty)\cap A$ are disjoint non-empty open sets different from $A$.

Now, since $M$ is connected and $f$ is continuous, $f(M)\subseteq I$ must be connected. On the other hand, if $f(M)$ contained at least two different points, then $f(M)$ would be disconnected, as we have seen. Hence $f$ is constant.

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