If there are e.g. $2$ sets $X,Y$ then the Cartesian product $X\times Y$
can be looked at as a set of ordered pairs: $\left\{ \left\langle x,y\right\rangle \mid x\in X\wedge y\in Y\right\} $.
But what if you want a notion of Cartesian product concerning a whole
bunch of sets? Let's say we have the sets $X_{i}$ where $i$ ranges
over indexset $I$. Then functions take the place of the ordered pairs.
Elements of $\prod_{i\in I}X_{i}$ are functions $f$ having $I$
as domain and with $f\left(i\right)\in X_{i}$ for each $i\in I$.
We want to give these functions a common codomain. Note that for every
$i\in I$ the set $X_{i}$ must be a subset of the codomain of these
functions leading to the set $\bigcup_{i\in I}X_{i}$ as codomain.
This together gives the definition that you mention in your question.
In the example that you mention: $\left\{ 1,2\right\} \times\left\{ 2,3\right\} =\left\{ \left\langle 1,2\right\rangle ,\left\langle 1,3\right\rangle ,\left\langle 2,2\right\rangle ,\left\langle 2,3\right\rangle \right\} $
we need an indexset $I$ that has exactly $2$ elements and we can
do it with $I=\left\{ 1,2\right\} $ as you propose.
Then e.g. element
$\left\langle 1,3\right\rangle $ corresponds with function $f:\left\{ 1,2\right\} \rightarrow\left\{ 1,2,3\right\} $
that is prescribed by $1\mapsto1$ and $2\mapsto3$.
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:
- Ordered sets
- Groups (additive, without $\mathsf{AC}$)
- Rings
- Groups (multiplicative, minus $0$)
- Fields (giving $\Bbb R\times\Bbb R$ the complex product)
- $\Bbb R$-Vector spaces
- Metric (or normed, or inner product) spaces
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.
Best Answer
Each of the sets $A$ and $B$ has a single element. For a moment call those elements simply $a$ and $b$ to avoid being distracted by the specific nature of the elements; then it should be clear that the only member of $A\times B$ is the ordered pair $\langle a,b\rangle$. Now let’s go back and recall what $a$ and $b$ actually are: $a=\{5\}$, and $b=\varnothing$. Thus, $A\times B$ contains just the ordered pair $\langle\{5\},\varnothing\rangle$:
$$A\times B=\{\langle\{5\},\varnothing\rangle\}\;.$$