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This question is inspired by Exercise 1.44 Mathematical Proofs, 2nd Ed, by Gary Chartrand et al. Given that $A = \{\emptyset, \color{green}{\{{\emptyset}\}}\}$, I understand that $A \times \mathcal{P}(A) = \LARGE{\{} \normalsize{ (\emptyset, \emptyset), (\emptyset, \color{green}{\{{\emptyset}\}}), (\emptyset, \color{blue}{\{\{\emptyset\}\}}), (\emptyset, A), (\color{green}{\{{\emptyset}\}},\emptyset), ( \color{green}{\{{\emptyset}\}}, \color{green}{\{{\emptyset}\}}), ( \color{wgreen}{\{{\emptyset}\}}, \color{blue}{\{\{\emptyset\}\}} ), (\color{green}{\{{\emptyset}\}} ,A) } \LARGE{\}} $.
I also understand that these 3 all differ: $\emptyset$ = an empty box, $\color{green}{\{{\emptyset}\} \text{= a box containing an empty box }}, \color{blue}{\{\{\emptyset\}\} \text{= a box containing a box containing an empty box} }.$

However, what is the meaning of an ordered pair containing any one of $\emptyset$, or $\color{green}{\{{\emptyset}\}}$ or $\color{blue}{\{\{\emptyset\}\}}$? As 3 examples, $(\emptyset, \emptyset), (\emptyset, \color{blue}{\{\{\emptyset\}\}}), ( \color{green}{\{{\emptyset}\}}, \color{blue}{\{\{\emptyset\}\}} )$ unhinge me because I only obtained them by the definition of Cartesian product; I am daunted and nescient about their true meaning. Could the 8 ordered pairs above be represented graphically? Or is there intuition or another interpretation?

I have referenced Ordered pairs in a power set.

$\large{\text{Supplement to Alex Mardikian and Professsor Scott's Answers :}}$

With many thanks to your answers, I now understand the ordered pairs can be bijected with natural numbers for simplicity.

Nonetheless, without rewriting or paring them, is it possible to understand directly the meaning of an ordered pair containing any one of $\emptyset$, or $\color{green}{\{{\emptyset}\}}$ or $\color{blue}{\{\{\emptyset\}\}} ?$ The bijections have taken away some of my "angst", but I still feel fazed by ordered pairs like $(\emptyset, \emptyset), (\emptyset, \color{blue}{\{\{\emptyset\}\}}), ( \color{green}{\{{\emptyset}\}}, \color{blue}{\{\{\emptyset\}\}} )$.

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I am appalled that this has four votes to close. A question asking for help with a mathematical intuition is obviously a serious question. – Brian M. Scott Jul 14 '13 at 5:19
This closing is a travesty. – Brian M. Scott Jul 14 '13 at 5:37
@BrianM.Scott: Thank you very much for your support, Professor. – Timere Jul 14 '13 at 6:05
Please use colors sparingly, if at all. – Cameron Buie Jul 15 '13 at 13:25
In regards to your supplement, I am not quite sure of how to think of the elements $\emptyset,\{\emptyset\},\{\{\emptyset\}\},\{\emptyset,\{\emptyset\}\}$ beyond just their uniqueness between each other. One thing about the only notable place I have seen these particular elements is when constructing the natural numbers starting with the empty set, where \begin{equation}0:=\emptyset\text{ and }n+1:=n\cup\{n\}. \end{equation}So starting with $1 = 0 + 1 = \emptyset \cup\{\emptyset\} = \{\emptyset\}$, and you just keep adding 1. Can't think of any more on the ordered pairs though, sorry. – Alex Jul 15 '13 at 14:10
up vote 4 down vote accepted

Here’s the set of ordered pairs $A\times\wp(A)$, with $A$ listed across the bottom and $\wp(A)$ along the lefthand edge:

$$\begin{array}{r|ll} \{\varnothing,\{\varnothing\}\}&\langle\varnothing,\{\varnothing,\{\varnothing\}\}\rangle&\langle\{\varnothing\},\{\varnothing,\{\varnothing\}\}\rangle\\ \{\{\varnothing\}\}&\langle\varnothing,\{\{\varnothing\}\}\rangle&\langle\{\varnothing\},\{\{\varnothing\}\}\rangle\\ \{\varnothing\}&\langle\varnothing,\{\varnothing\}\rangle&\langle\{\varnothing\},\{\varnothing\}\rangle\\ \varnothing&\langle\varnothing,\varnothing\rangle&\langle\{\varnothing\},\varnothing\rangle\\ \hline &\varnothing&\{\varnothing\} \end{array}\tag{1}$$

Here’s the same table of $B\times\wp(B)$, where $B=\{0,1\}$:

$$\begin{array}{r|ll} \{0,1\}&\langle0,\{0,1\}\rangle&\langle1,\{0,1\}\rangle\\ \{1\}&\langle0,\{1\}\rangle&\langle1,\{1\}\rangle\\ \{0\}&\langle0,\{0\}\rangle&\langle1,\{0\}\rangle\\ \varnothing&\langle0,\varnothing\rangle&\langle1,\varnothing\rangle\\ \hline &0&1 \end{array}\tag{2}$$

As you can see, they have exactly the same structure: only the labels are different. If you go through $(1)$ replacing $\varnothing$ by $0$ and $\{\varnothing\}$ by $1$, you get exactly $(2)$.

And here, to make the structure even clearer, is $B\times C$, where $C=\{0,1,2,3\}$

$$\begin{array}{r|ll} 3&\langle0,3\rangle&\langle1,3\rangle\\ 2&\langle0,2\rangle&\langle1,2\rangle\\ 1&\langle0,1\rangle&\langle1,1\rangle\\ 0&\langle0,0\rangle&\langle1,0\rangle\\ \hline &0&1 \end{array}\tag{3}$$

If in $(2)$ you replace every second coordinate $\varnothing$ by $0$, every second coordinate $\{0\}$ by $1$, every second coordinate $\{1\}$ by $2$, and every second coordinate $\{0,1\}$ by $3$, you get exactly $(3)$. The three Cartesian products $A\times\wp(A)$, $B\times\wp(B)$, and $B\times C$ have the same basic structure. This structure is easiest to see in $(3)$, but that’s only because the notation is less cluttered.

Added: I think that the problem that you’re having isn’t really with the ordered pairs themselves, but rather with the objects appearing in them. In practical terms an ordered pair is just a list of two things arranged so that there is a first thing and a second thing (which may be the same thing listed twice), and we can tell which is which. As far as the ordered pair aspect is concerned, there’s no difference between $\langle\varnothing,\{\{\varnothing\}\}\rangle$ and $\langle 3,7\rangle$: both are ordered pairs whose first and second components are unequal. Each becomes a different ordered pair if you reverse the order of its components: $\langle\{\{\varnothing\}\},\varnothing\rangle\ne\langle\varnothing,\{\{\varnothing\}\}\rangle$, and $\langle 7,3\rangle\ne\langle3,7\rangle$.

To put it a little differently, the meaning of $$\langle\text{thing}_1,\text{thing}_2\rangle$$ is simply list of two things, the first one being $\text{thing}_1$ and the second $\text{thing}_2$. The identities of $\text{thing}_1$ and $\text{thing}_2$ have no bearing on this meaning. We might say that they’re buried one layer deeper: after identifying $\langle\text{thing}_1,\text{thing}_2\rangle$ as an ordered pair of entities, we can worry about just what those entities are. If $\text{thing}_1=3$ and $\text{thing}_2=7$, and the context is plotting a point in the Cartesian plane, we hardly notice this step, because we’re very familiar with $3$ and $7$ as real numbers. If $\text{thing}_1=3$ and $\text{thing}_2=7$, and the context is a first serious course in set theory, we might have to think a bit harder about this step, because in that context it’s likely that $3=\{0,1,2\}$ and $7=\{0,1,2,3,4,5,6\}$, so that


But the added complication has nothing to do with the ordered pair structure: it’s all in the two objects that are the components of the ordered pair.

The same is true if $\text{thing}_1=\varnothing$ and $\text{thing}_2=\{\{\varnothing\}\}$: $\langle\varnothing,\{\{\varnothing\}\}\rangle$ is just a list whose first element is $\varnothing$, and whose second element is $\{\{\varnothing\}\}$. If you don’t feel that you have an intuitive grip on this, the problem is unlikely to be in the ordered pair structure itself, in the notion of a first thing and a second thing; it’s much likelier to result from the relative unfamiliarity of the things themselves. That’s a problem that to a large extent will solve itself over time, provided that you keep working with the concepts. In the short term it may help to make a conscious effort to think about ordered pairs in ‘layers’:

Okay, this is an ordered pair. Its first component is $\text{thing}_1$, and its second component is $\text{thing}_2$. It’s an element of the function $f$, so I know that $f(\text{thing}_1)=\text{thing}_2$. Now what are these things? Well, $\text{thing}_1=\varnothing$ and $\text{thing}_1=\{\{\varnothing\}\}$, so $f(\varnothing)=\{\{\varnothing\}\}$. Okay: the function $f$ assigns to the empty set the set $\{\{\varnothing\}\}$. Do I really need to know more than that right now?

In this example the ordered pair is in a middle layer, with the function $f$ ‘above’ it and the objects $\text{thing}_1$ and $\text{thing}_2$ ‘below’ it. It’s part of the internal structure of $f$, and $\text{thing}_1$ and $\text{thing}_2$ are part of its internal structure. They in turn may have internal structure; in this case $\varnothing$ has none, but $\{\{\varnothing\}\}$ clearly does have some. The details of that internal structure may or may not matter. If they do, I’ll have to think about them: $\{\{\varnothing\}\}$ is the set whose only element is $\{\varnothing\}$, which in turn is the set whose only element is $\varnothing$. If not, I can just treat $\{\{\varnothing\}\}$ as a fancy label for some set whose precise nature isn’t important, at least at the moment. If it becomes important later, I can worry about it then.

Actually, this advice applies to all mathematical notation. You don’t have to grasp a complicated expression all at once: it’s fine to build up an understanding of it a piece at a time. An equation, for instance, by definition has the form $A=B$. We see that, and we immediately know the type of expression with which we’re dealing. Now what are $A$ and $B$? And we go on from there, making sense of $A$ and $B$.

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Thank you very much for your continued help and helpful response, Professor. Could you please look at a supplement in my original post? – Timere Jul 15 '13 at 13:38
@LePressentiment: I’ve added a longish discussion that may (or may not) be of some use. – Brian M. Scott Jul 15 '13 at 20:08
Thank you very much for your detailed addition. It was absolutely helpful! I really enjoyed and will keep in mind your step-by-step illustration of a mathematician's stream of thinking and your general advice in the end. They seem rare in textbooks, yet I find them invaluable. – Timere Jul 16 '13 at 15:02
@LePressentiment: You’re very welcome. – Brian M. Scott Jul 17 '13 at 20:39

When you say that you are daunted by the appearance of any of these 3 sets in an ordered pair, I think you really mean elements of the sets, where $\emptyset$ and $\{\emptyset\}$ are elements of $A$ and $\emptyset$, $\{\emptyset\}$, $\{\{\emptyset\}\}$, and $A$ are elements of $\mathcal{P}(A)$.

Think of the ordered pairs just like you would points on a graph, and imagine that for example we have the following numerical assignment to the elements of your sets respectively: \begin{equation} \emptyset : 1 \qquad \{\emptyset\} : 2 \qquad \{\{\emptyset\}\} : 3 \qquad A = \{\emptyset,\{\emptyset\} \} : 4 \end{equation} The reason I am suggesting this is because as each of those 3 elements are unique, you want to think of them as such. Now by taking the ordered pairs you have and replacing the element with this numerical assignment, you can have a visual representation on a graph of how to uniquely identify each ordered pair. So for $(\emptyset,\{\emptyset\})$ you can now imagine $(1,2)$, and can note the difference between a different ordered pair $(2,1)$ which would then be $(\{\emptyset\},\emptyset)$.

If you take the sets $\mathbb{R} \times \mathbb{R}$, where $\mathbb{R}$ is the set of all real numbers, you do in fact get the Cartesian coordinate system as a visual representation of the ordered pairs naturally. So if you take this into consideration, your intuition of seeing these pairs as points on a graph do make sense as long as you know which elements are unique.

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Thank you very much for your helpful response. Could you please look at a supplement in my original post? – Timere Jul 15 '13 at 13:38

If you understand a set as a box, you may understand an ordered pair as a special kind of box with two (distinguishable) drawers, say one blue drawer and one red drawer.

Then, for example, with $(\emptyset,\{\emptyset\})$ you should see a box with two drawers, when you open the blue one, you find an empty box, and when you open the red one, you find a box containing an empty box.

And if you don't like special boxes with coloured drawers, you can "encode" those into standard boxes by defining for example $(a,b)= \{\{a\};\{a;b\}\}$. The only point to this encoding is not having to use strange drawers, and still have that $(a,b) = (c,d) \iff a=c \land b=d$ (whose proof is spectacularly boring). The encoding is absolutely useless when you have to reason about ordered pairs.

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Thank you very much for your analogies. I've upvoted but would still have liked to choose multiple answers. – Timere Jul 16 '13 at 15:03

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