Cardinalities of $\mathbb{R^{2}}$ and $\mathbb{C}$ and isomorphisms 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.
 A: 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)
A: 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.
