# What's the Cartesian Product of $\{x\}$ and $\{x\}$?

The exercise I'm working on is "Show that $\{x\} \times \{x\} = \{\{\{x\}\}\}$".

The solution I'm seeing says that this is equal to $\{\{\{x\}\}\}$, but I don't get this yet.

Here's what I do understand:

• $\{x\}\times\{x\}$ should be the set of ordered pairs whose first coordinate is in $\{x\}$ and whose second coordinate is also in $\{x\}$.
• The only thing in $\{x\}$ is $x$.
• The cartesian product is a set.
• An ordered pair is a set.

So if $x$ is the only thing in $\{x\}$, then the only ordered pair of $\{x\} \times \{x\}$ should be $\{x\}$, and the cartesian product should be the set of all of those ordered pairs, i.e. $\{\{x\}\}$.

Any hints as to what I'm missing?

• Note that this is a consequence of a standard way of defining an ordered pair. – copper.hat Nov 27 '17 at 4:10
• It's so interesting to me that an ordered pair is defined -- in this standard way -- as a set of sets. I wonder if that's conceptually or even psychologically true, or just a formal trick. Meaning, is our concept of an ordered pair really of a collection of a collection? I'm confused thinking about this. – MBP Nov 27 '17 at 4:13
• There are various definitions, see en.wikipedia.org/wiki/…. – copper.hat Nov 27 '17 at 4:29
• @MBP Everything in typical set theory (i.e. ZFC) is a set and the only way to get any structure at all is to nest them. The only thing that isn't a set of sets is the empty set (and arguably that's just an empty set of sets...) I would say our concept of an ordered pair isn't a set of sets, but the point of set theory is to show that it (and every other mathematical concept) can be implemented as a set of sets, but no particular implementation is the concept. – Derek Elkins Nov 27 '17 at 5:35
• – user236182 Nov 27 '17 at 15:51

The set $\{x\}\times \{x\}$ is the singleton containing the ordered pair $(x,x)$, i.e. $\{x\}\times\{x\} = \{(x,x)\}$. By definition, $(x,x) = \{\{x\},\{x,x\}\}$. This last set is equal to $\{\{x\},\{x\}\}$ since $\{x,x\} = \{x\}$, and this final set is equal to $\{\{x\}\}$ since $\{x\} = \{x\}$. Hence $\{x\}\times\{x\} = \{\{\{x\}\}\}$.