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Determine if the following sets are countable or not:

(1) The set of functions $\mathbb{R}\to\mathbb{R}$ with values in $\mathbb{Z}$

(2) The set of continuous functions $\mathbb{R}\to\mathbb{R}$ with values in $\mathbb{Z}$

My guess is that (1) is not countable and I think the reason is the following: I know that the set of all functions $\mathbb{N}\to \{0,1\}$ is not countable. Since the cardinality of all functions $\mathbb{N}\to \{0,1\}$ is less or equal than the cardinality of all functions $\mathbb{R}\to\mathbb{R}$ with values in $\mathbb{Z}$, the latter set is not countable. Is it correct?

I have no idea how to do (2).. I appreciate any help and hint. Thank you

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    $\begingroup$ You know the Intermediate Value Theorem? $\endgroup$ – Lord Shark the Unknown Jan 18 '18 at 19:47
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    $\begingroup$ What do you mean "with values in $\mathbb Z$"? DO you m fnction $\mathbb R \to \mathbb Z$? Why not right it as such. The way you stated your reasoning for 1) isn't yet correct but on the right track. $\{f:\mathbb N\to \{0,1\}\} \subset \{f:\mathbb N\to \mathbb Z\} \subset \{f:\mathbb R\to \mathbb Z\}$. And now you can state it. 2) has a trick in that continuous function can only be constant. $\endgroup$ – fleablood Jan 18 '18 at 19:53
  • $\begingroup$ thank you both. @LordSharktheUnknown yes I know this theorem $\endgroup$ – user472520 Jan 18 '18 at 19:57
  • $\begingroup$ ah, I can use the intermediate value theorem to see that continuous functions $\mathbb{R}\to\mathbb{Z}$ must be constant, right $\endgroup$ – user472520 Jan 18 '18 at 20:09
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Hint for $(1)$:

Is the set of functions $\Bbb N\to\{0,1\}$ countable? What do you think about $\Bbb R\to\Bbb Z$?

Hint for $(2)$:

A continuous function $\Bbb R\to \Bbb Z$ is constant. How many integer constants are there?

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  • $\begingroup$ Thank you. I have already a solution for (1). Ok, I wasn't aware of that continuous functions from the reals mapping into the integers must be constant. $\endgroup$ – user472520 Jan 18 '18 at 19:55
  • $\begingroup$ Then, the cardinality of continuous functions $\mathbb{R}\to\mathbb{Z}$ agrees with the cardinality of $\mathbb{Z}$, and is therefore countable. Is it correct? $\endgroup$ – user472520 Jan 18 '18 at 20:00

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