Does it make sense to say that $\{1,4\}$ is a subset of $(\{1,2\},\{4,6\})$?

Say we have $X=\{1,2,3\}$, $Y=\{4,5,6\}$ and take the cartesian product of their power sets $A=\mathcal{P}(X)\times \mathcal{P}(Y)$. Now, $(\{1,2\},\{4,6\})\in A$, although what are its subsets of this element? Is $\{1,4\}$ a subset of $A$, or is $\{1,2\}$?

It feels like $\{1,2\}$ should be a subset of $A$, although it wouldn't actually be in $A$, would it? $\{1,4\}$, on the other hand, would be an element of $A$.

Or does $(\{1,2\},\{4,6\})$ just have no proper subsets?

• $(\{1,2\},\emptyset)$ is an element of $A$ but not $\{1,2\}$ which is an element of $\mathcal P(X)$ and therefore a subset of $X$. – Surb Dec 10 '15 at 20:09
• "...what are its subsets?" This can be interpreted as "...what are the subsets of $(\{1,2\},\{4,6\})$?" or "...what are the subsets of $A$?" The next sentence suggests that you intended the second interpretation; if so, perhaps you could make this explicit. – TonyK Dec 10 '15 at 20:15
• Thanks for pointing that out - I've edited the question so it's clearer – man_in_green_shirt Dec 10 '15 at 20:18
• In that case, the subsets of an ordered pair depend on your definition of ordered pair. Are you using Kuratowski's definition? – TonyK Dec 10 '15 at 20:22
• I just noticed I accidentally downvoted your question! Sorry about that. If it gets edited, I will be able to remove my downvote. – TonyK Dec 10 '15 at 20:57

(answer modified after question was edited)

$A$ and its subsets consist of ordered pairs of sets, not individual numbers, sets of numbers or sets of ordered pairs.

$(\{1,2\},\{4,6\})\in A.$ It is an ordered pair but not a set of ordered pairs. It is not a set, unless you are using a specific model for ordered pairs (which is sometimes done in logic handbooks where all objects must be sets).

Subsets are sets that contain zero or more elements from the given set and nothing else. So a subset of any subset of $A$ would automatically have to consist of ordered pairs.

• Yes, my bad, got the brackets the wrong way round. I'll edit the question now – man_in_green_shirt Dec 10 '15 at 19:58

This may have to do with the quirky way of defining the ordered pair $(a,b)$ as the set $\{\{a\}, \{a,b\}\}$. Thus, $(\{1,2\}, \{4,6\} ) = \{ \{\{1,2\}\}, \{ \{1,2\}, \{4,6\} \} \}$. Is $\{1,4\}$ a subset of that ? Not really. Can $\{1,4\}$ be a subset of some $(a,b) = \{ \{a\}, \{a,b\} \}$ ? Not really. So it seems the question is valid and has a negative answer.

• Thanks for your answer. So essentially, no element of $A$ can have a proper subset, or be a proper subset of any other element of $A$, right? – man_in_green_shirt Dec 10 '15 at 20:34
• Just to clarify orangeskid's answer: $\{1,4\}$ is definitely not a subset of $\{ \{\{1,2\}\}, \{ \{1,2\}, \{4,6\} \} \}$. – TonyK Dec 10 '15 at 20:39
• $(\{1,2\},\{4,6\})$ has two elements under the Kuratowski definition, and thus has four subsets. @man_in_green_shirt – Akiva Weinberger Dec 10 '15 at 20:42
• Ok. But apart from the empty set and the last one, are those $4$ subsets elements of $A$? – man_in_green_shirt Dec 10 '15 at 21:19

Edited to add: Now that I've bullied man_in_green_shirt into choosing a definition of ordered pair, we can get to grips with this.

If an ordered pair $(a,b)$ is defined as $\{\{a\},\{a,b\}\}$, then we see that it has two elements (assuming that $a$ and $b$ are distinct), and therefore four subsets: $\emptyset, \{\{a\}\}, \{\{a,b\}\},$ and $\{\{a\},\{a,b\}\}$. Hence the four subsets of $(\{1,2\},\{4,6\})$ are:

$\emptyset$
$\{\{\{1,2\}\}\}$
$\{\{\{1,2\},\{4,6\}\}\}$
$\{\{\{1,2\}\},\{\{1,2\},\{4,6\}\}\}$

I hope I have that right. It looks gruesome in MathJax.

Original answer before the question was clarified:

The elements of $A$ are ordered pairs $(S_X,S_Y)$, where $S_X \subset X$ and $S_Y \subset Y$. So if $S_A$ is a subset of $A$, then the elements of $S_A$ are ordered pairs too. In particular, neither $\{1,4\}$ nor $\{1,2\}$ is a subset of $A$. Here is a typical subset of $A$:

$$\{(\{1,2\},\{4,6\}),(\emptyset,\{4,5,6\}),(\{2\},\{4,6\}),(\{1,2,3\},\{6\}),(\{1,3\},\{5\})\}$$