Using the tools of linear-algebra it seems like $\mathbb{R^{2}}$ or $\mathbb{R}$ $\times$ $\mathbb{R}$ is isomorphic to $\mathbb{C}$, since both of the spaces are of dimension 2. Does this mean that the these sets are equinumerous? $\mathbb{R}$ is a subspace of $\mathbb{C}$, but I am wondering how could one prove that they are equinumerous without using isomorphisms (assuming they imply equinumerosity); would this require cardinal arithmetic? Thanks.
Cardinalities of $\mathbb{R^{2}}$ and $\mathbb{C}$ and isomorphisms
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0Isomorphic as what? Real vector spaces? Yes, they both have dimension $2$. – 2011-03-18
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0@Qiaochu: Yes as real vector spaces (only real scalar multiplication allowed), does that imply that the sets are equinumerous? Can this be proven set-theoretically without isomorphisms (assuming isomorphisms imply equinumerosity). – 2011-03-18
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2Yes, of course $\mathbb{R}^2$ is equinumerous with $\mathbb{C}$: just take the bijection $(x,y) \mapsto x+yi$. In fact, $\mathbb{C}$ is often defined to be the set $\mathbb{R}^2$ with appropriately-defined addition and multiplication. – 2011-03-18
2 Answers
We define cardinality as equivalence class under the relation "There exists a function which is 1-1 and onto between $A$ and $B$", in which case we say that $A$ and $B$ have the same cardinality.
Since the mapping $(x,y) \mapsto x+iy$ is a bijection it is clear that $|\mathbb{R}^2|=|\mathbb{C}|$.
In the more general case, the function need not be a linear isomorphism, and it requires only the properties I mentioned above. This is because there is an essential lack of structure in set theory, compared to linear algebra (for example) in the sense that functions don't require to preserve addition and scalar multiplication.
This can also be proved by cardinal arithmetics, but it is just as immediate (due to the very obvious bijection I wrote above)
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2That is actually _not_ how we define cardinality because the class of sets is not a set, so one can't speak of equivalence relations on it in the usual way. – 2011-03-18
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3@Qiaochu: This definition can be fixed by a simple application of [Scott's trick](http://en.wikipedia.org/wiki/Scott%27s_trick), which gives a uniform way of picking representative subsets of equivalence classes which are proper classes. – 2011-03-18
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2@Chris: right, but it's not necessary to have a definition of cardinality to answer this question; all we need to know is what a bijection is. – 2011-03-18
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4Qiaochu: If one wants to be 100% accurate then you are correct, but by using Scott's trick this is easily fixable. I was taught that sometimes hiding some of the formal backbone can really clarify the overall idea. I sense that in this case my approach was likely favourable in comparison of details and 100% formalism. – 2011-03-18
While it's true that $\mathbb{R}^2$ and $\mathbb{C}$ are equinumerous, it's misleading to say that it has anything to do with their both having dimension 2; indeed, $\mathbb{R}^2$ and $\mathbb{R}$ are also of equal cardinalities, and in fact so are $\mathbb{R}$ and $\mathbb{R}^n$ for any $n$; all of these sets have cardinality $\mathfrak{c}$. You don't need an isomorphism to show this (after all, $\mathbb{R}^n$ and $\mathbb{R}^m$ aren't isomorphic as vector spaces if $m\not= n$, but they still have the same cardinality), but you do need a bijection - that's the very definition of equinumerous, after all. Once you have a bijection between $\mathbb{R}^2$ and $\mathbb{R}$ you can use this to build up all of the other bijections by just composing it with itself; finding this bijection is a nice, classic exercise, and I highly recommend it.
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0Thanks a lot, Steven. So am I correct in my understanding that isomorphism always implies equinumerosity, while equinumerosity does not always imply isomorphism (which makes sense due to the additional conditions placed on the isomorphisms)? – 2011-03-18
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0Well, you can't say 'isomorphic' without saying what structure you're treating your isomorphism with respect to - after all, two objects being of the same cardinality _is_ an isomorphism with respect to the 'no additional structure' of Set. But yes, it's basically correct to say that isomorphism (over some structure that extends Set) implies equinumerosity, because you can always forget the extra structure. – 2011-03-19