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I am having trouble understanding this - especially the arbitrary part.

$X$ is an arbitrary set, $M\subseteq X$ is an arbitrary denumerable set, $M'\subseteq X$ is an arbitrary finite set.

1) Is $M\times M$ denumerable?

2) Is $M'\times M'$ denumerable?

What does arbitrary mean in this context? I am having trouble understanding the difference between $M$ and $M'$. I do not know how to work with an "arbitrary set". What is that?!

Andrés E. Caicedo
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four-eyes
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  • "Arbitrary" has no formal effect. See here. – hmakholm left over Monica Nov 06 '15 at 11:57
  • $M$ and $M'$ are just two sets - denumerable and finite, respectively. "Arbitrary" just means that we let them be any such sets, i.e., we don't assume that $M\supseteq M'$ for example. – Noah Schweber Nov 06 '15 at 11:59
  • @NoahSchweber Thanks. That helped. So both sets must be denumberable? – four-eyes Nov 06 '15 at 12:05
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    @Chrissl: It says explicitly that $M'$ must be finite. Usually "denumerable" means "countably infinite", so $M'$ cannot be denumerable. – hmakholm left over Monica Nov 06 '15 at 12:07
  • @HenningMakholm But why are infite sets denumerable and finite sets not?! If something is denumerable in infinity, something finite should be denumerable too?! – four-eyes Nov 06 '15 at 12:10
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    @Chrissl: Not all infinite sets are denumerable. At this point you really should look in your textbook for its exact definition of the word "denumerable". – hmakholm left over Monica Nov 06 '15 at 12:11
  • @HenningMakholm Of course not all infinite sets are denumerable. I am aware of that! What I cannot grasp is why something infinte is not denumerable. – four-eyes Nov 06 '15 at 12:12
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    @Chrissl: I don't see any difference between what you claim you are "of course" aware of, and the thing you "cannot grasp". Which difference is there between them? – hmakholm left over Monica Nov 06 '15 at 12:12
  • @HenningMakholm I understand that infinite(!) sets can be either denumerable or nondenumerable. What I do not understand is, why M' is nondenumerable. That is a finite set. If something is finite, I should be able to denumerable it?! – four-eyes Nov 06 '15 at 12:15
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    @Chrissl: Again, you need to tell us the exact definition of the word "denumerable" you're working with here. – hmakholm left over Monica Nov 06 '15 at 12:15
  • @HenningMakholm Like in Cantor. – four-eyes Nov 06 '15 at 12:16
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    @Chrissl: "Like in Cantor" is not a definition of the word "denumerable". If you refuse to reveal the actual definition you're working with, we won't be able to help you. – hmakholm left over Monica Nov 06 '15 at 12:17
  • @Chrissl Some people use "denumerable" to mean "countable and infinite," others just use it to mean "countable (possibly finite)". So you need to clarify. – Noah Schweber Nov 06 '15 at 12:17
  • @HenningMakholm I meant with "like in Cantor" the way he uses denumerable when he proves that |N| = |Q+| or that one https://en.wikipedia.org/wiki/Cantor's_diagonal_argument. Basically, denumerable means (for me) bijection https://en.wikipedia.org/wiki/Bijection. – four-eyes Nov 06 '15 at 12:20
  • @Chrissl: That is still not an actual definition of the word "denumerable". If you want help, provide an actual verbatim quote of the definition in the textbook you're using. (And nobody uses the word "denumerable" as a synonym of "bijection"). – hmakholm left over Monica Nov 06 '15 at 12:21
  • @HenningMakholm A set is denumerable if there a bijection. – four-eyes Nov 06 '15 at 12:22
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    @Chrissl: You're missing several words there. You seem to be deliberately difficult here, repeatedly refusing to quote the definition you're using. I conclude you're just a troll, not really interested in being helped, and I have better things to do than keep trying to get you to reveal the definitions you're ostensibly asking for help with understanding. Goodbye! – hmakholm left over Monica Nov 06 '15 at 12:25
  • @HenningMakholm Thanks for sticking with me :) Have a good one! – four-eyes Nov 06 '15 at 12:27

1 Answers1

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The product of two finite sets is finite and the product of two denumerable sets is denumerable. This is pretty easy to prove. To prove the latter statement you need to find a bijection from the set of positive integers onto the set of ordered pairs of positive integers. To do this all you need to do is find a way of putting all the ordered pairs of positive integers into an infinite sequence indexed by the positive integers. If you write them up in an infinite two-dimensional array then you can actually draw a picture of how it works. Have a think about it.

Rupert
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  • I do know how to prove it. But I do not know how to work with an "arbitrary set". What is that?! – four-eyes Nov 06 '15 at 11:59
  • @Chrissl: Have you followed the link I keep giving you? – hmakholm left over Monica Nov 06 '15 at 12:01
  • @HenningMakholm Yep. http://math.stackexchange.com/questions/308516/terminology-arbitrary-vs-finite/308562#308562 But I did not understand your answere :) – four-eyes Nov 06 '15 at 12:02
  • "Arbitrary subset" just means "any possible subset". So the Reader is asked to reason about a general subset (using only the assumed finiteness or denumerability), and not to provide an argument that works only for a particular subset. – hardmath Nov 08 '15 at 16:08