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How do I prove that the following sets have equal cardinality?

  1. $|\Bbb N \times\Bbb N \times\Bbb N| = |\Bbb N|$ ($|\Bbb N \times\Bbb N| = |\Bbb N|$ also for that matter)
  2. $|\Bbb Z \times\Bbb Z| = |\Bbb Z|$
  3. $|\Bbb R \times\Bbb R| = |\Bbb R|$

Thank you!

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What did you try? Did you try searching the site? – Asaf Karagila Jan 5 '13 at 18:19
yes I tried searching the site – Yechiel Labunskiy Jan 5 '13 at 18:26
Look here, here, here, here, and here; among them they answer all of your questions. – Brian M. Scott Jan 5 '13 at 18:41
Thank you Brian! Guess I didn't know what to search for. – Yechiel Labunskiy Jan 5 '13 at 18:54
Wie ought to have something like an abstract duplicate "When is $|X\times X|=|X|$?" – Hagen von Eitzen Jan 6 '13 at 10:34

It is easy to produce a sequence that includes all the elements in $\Bbb N\times\Bbb N\times\Bbb N$.

E.g. $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, $(0,0,1)$, $(2,0,0)$, $(1,1,0)$, $(1,0,1)$, $(0,2,0)$, $(0,1,1)$, $(0,0,2)$, $(3,0,0)$, $(2,1,0)$, $(2,0,1)$,...

This can be adjusted to the case of $\Bbb Z\times\Bbb Z\times\Bbb Z$ and/or any finite number of copies of $\Bbb N$ or $\Bbb Z$, or even $\Bbb Q$.

To show that $|\Bbb R|=|\Bbb R\times\Bbb R|$ requires, of course, a different argument.

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How do I map that sequence to |N| with a one-to-one and an unto function? – Yechiel Labunskiy Jan 5 '13 at 18:55
Well, you map $1$ to the first element, $2$ to the second element, $3$ to the third element and so on. Recall that a sequence in a set $X$ is just a function $\Bbb N\rightarrow X$. If a sequence contains exactly once all the elements of $X$, then the sequence, thought as a function, is bijective. – Andrea Mori Jan 5 '13 at 19:12

Note that having size $|N|$ means something is infinite but enumerable; meaning, roughly, that you can come up with a way of listing all the elements in a single row.

Note: I am abiding by $N = \{0, 1, 2, 3, \ldots\}$.

To show $|N \times N| = |N|$, you could start by listing all the elements $(m, n)$ with $m,n \in N$ such that $m + n \leq 0$. Next, list all the elements with $m+n \leq 1$, and so forth. In this way, you produce a list of the elements; a list of the form:

$a_0, a_1, a_2, a_3 \ldots$ which can be matched up with elements of $N$ using $a_n \rightarrow n$.

The case for $N \times N \times N$ is similar, as is the case for anything $Z$ related.

Your third problem is a bit more involved, but one place to start is by thinking about decimal expansions. I won't say anything more unless you post what you have tried thus far.

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Pairs like (2,2) and (3,1) will both map to 4 which means it's not one-to-one, am I missing something? Don't really know where to start on R x R. Thanks – Yechiel Labunskiy Jan 5 '13 at 18:29
The key is to come up with an organized way of listing all of them. One way to avoid the sort of problem you have described is to always list $(a,b)$ before $(c,d)$ when $a + b = c + d$ and $a < c$. So the list would begin: $(0, 0), (0, 1), (1, 0), (0, 2), (1, 1), (2, 0), (0, 3), (1, 2), (2, 1), (3, 0), (0, 4), (1, 3), (2, 2), (3, 1)$ Of course, a similar approach can be used to demonstrate that $\mathbb{Q}$ is countable. – Benjamin Dickman Jan 6 '13 at 10:22

The function:$$f:\mathbb{Z}^+\times\mathbb{Z}^+\times\mathbb{Z}^+\rightarrow \mathbb{Z}^+$$ that sends $(i,j,k)$ to $(2(2j-1)2^{i-1}-1)2^{k-1}$ is a bijection.

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