Where the Cartesian Product of two sets $\mathbb A$ and $\mathbb B$ is such that $\mathbb A\times \mathbb B=\{{ (a,b)|a \in \mathbb{A}, b \in \mathbb{B}\}}$
In trying to understand the proof that $|\mathbb R\times \mathbb R|=|\mathbb R|$, the only part I can't understand is the use of the Cartesian Product. How can an element in the set $\mathbb R\times \mathbb R$ such as (0.7777777777……, 0.033333333333……) create the number (0.707373737373……) in $\mathbb R$?
The above step is necessary to invoke an injection but not a surjection from $\mathbb R\times \mathbb R$ to $\mathbb R$, since the proof I've been reading makes use of the Cantor-Bernstein-Shroeder theorem.
I can see that this number comes from adding the corresponding decimal digits from each of the numbers in (0.7777777777……….,0.033333333333…………), but since when is combining decimal digits in this way an "allowable" operation?
Basically what's bothering me is how an elemental pair can become a single element/number? That is to say – all I'm asking here is why $\\(a,b)$ can become a single entity say $k$ by the Cartesian product?
If you would please be kind enough to explain this to me I would be most grateful.
Thank You.
Best Answer
First, a bit of pedantry (sorry, this isn't related to what you ask but somebody had to say it). You write later in the question that the set $\{\Bbb R\times \Bbb R\}$ is mapped to the set $\{\Bbb R\}$. But these are sets of one element, and those elements are the sets $\Bbb R \times\Bbb R$ and $\Bbb R $, respectively. Those two sets (the curly bracket ones) are in bijection (in fact there's only one possible function between them), but it's clearly not the bijection you want. You want a bijection between the sets $\Bbb R\times \Bbb R$ and $\Bbb R$.
Anyway, regarding the original question, there's no problem with mapping a pair of real numbers to another real number, just think of any real function of two variables (e.g. $\tan xy$). I will admit that it can be surprising that this can be done bijectively, but once the bijection is there and we've shown it's well defined, it's no different than any other mapping.
If it helps, to understand set theoretical arguments, we can forget absolutely every and any algebraic, analytic, geometric, or what-have-you properties of the set in question. In most cases, yes, any algebraic (and other) structures will be destroyed by such bijections. We needn't worry, all that matters here is the well-definedness and the bijective property.
What might help even further, is to consider sets with no additional structure defined whatsoever, such as $\{\text{yum},\ssi,@,\text{top_hat_73}\}$. If you understand maps between these sets, then you can picture the elements of $\Bbb R$ as meaningless labels just like in the previous set, but there's a certain infinite amount of labels. In fact, we could place the string of characters "$\text{yum}$" after the decimal point of each real number, and still have the bijection in question.
Regarding the recent comment, the pair $(a,b)$ is a single entity to begin with, namely an element of the set $\Bbb R \times \Bbb R$. So, it doesn't become a single entity under the unnamed bijection you refer to, it is one that is simply associated to another entity, one of $\Bbb R$. Elements of sets are just elements of sets, no matter how the set was constructed (and there are far more atrocious constructions than the cartesian product).
I may have localized a source of confusion. You correctly intuit that $\Bbb R \times \Bbb R$ and $\Bbb R$ are not "structurally equivalent" in any of the usual ways. For instance, they are not isomorphic as:
And there's probably more. But they are in bijection, and the key is that the bijection can stomp all over the above properties if it wishes (and it does).
Maybe I can help you more if you describe the proof you've seen.