Is $A\times A^2=A^3$?

Question

If we were given a set $A=\{a,b\}$, then we have $A\times A=A^2=\{(a,a),(a,b),(b,a),(b,b)\}$ and $A^3=\{(a,a,a),(a,a,b),(a,b,a),(b,a,a),(a,b,b),(b,a,b),(b,b,a),(b,b,b)\}$.

Now my question is, what is $A\times A^2$ equal to? Is it $A^3$ or will the elements in this cartesian product be of the form $(x,(y,z))$, where $x\in A$ and $(y,z)\in A^2$? (I just used $x,y,z$ as arbitrary elements of $A$, just to generalize)

I should also note that I'm using Kuratowski's definition of ordered pairs and n-tuples.

Answer 1

In set theory, everything is a set¹, even ordered pairs/tuples, although the notion of an ordered pair predates formal set theory by at least a century². When we were developing a formal theory of sets, there needed to be a way we could express an ordered pair (and ordered triple) using only sets.

Mathematicians have, over the years, developed many ways to encode the idea of an ordered pair using set theory. The one that has gained the most currency is Kuratowski's definition:

$$(a, b) := \{\{a\}, \{a, b\}\}.$$

It has the key, defining property that $(a, b) = (c, d) \iff a = c \land b = d$. But, it's far from the only way of accomplishing this! As the link shows, there are other perfectly adequate ways of accomplishing this with sets.

What about ordered triples? Again, for us to use them in set theory, there needs to be some kind of definition in terms of sets.³ Like with ordered pairs, there are plenty of ways this could be accomplished, and I don't think there's a consensus on the best way to do this⁴. A simple way of doing this is to define $$(a, b, c) := ((a, b), c),$$ where ordered pairs are defined by whatever definition you prefer. Note that, if we take this to be the definition, then $A^2 \times A = A^3$, but $A \times A^2 \neq A^3$ (according to how you've interpreted $A^3$). This would be an example of a junk theorem.

We could also interpret $A^3$, not as a set of triples, but as a set of functions. Recall that $A^B$ is the set of functions from $A$ to $B$. Thus, $A^3$ could be interpreted as the set of functions from $A$ to the set $3$. Yes, $3$ is a set, like everything else covered by set theory. Without going into too many details, it is the set $\{0, 1, 2\}$ (where $0, 1, 2$ are also sets), so if we were to take $A = \{a, b\}$ say, then the following would be an element of $A^3$: $$\{(0, a), (1, a), (2, b)\}.$$ This would be another way of defining triples: $(a, b, c)$ could simply be the function $\{(0, a), (1, b), (2, c)\}$⁵. If we defined triples like so, then the set of triples $A^3$ agrees with the set of functions $A^3$. However, it means that $A^2 \times A$ and $A \times A^2$ are both different from $A^3$. Once again, this would be a junk theorem.

Ultimately, it doesn't really matter. Essentially, $A^2 \times A$ and $A^3$ (and $A \times A^2$) are the same thing. No mathematician or layperson really cares whether they are technically equal, or how you choose to encode your ordered pairs/triples. The two sets, regardless of how you sensibly define them, will have a very natural correspondence between the elements. For all intents and purposes, they are the same.

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¹ Set theory, as the name suggests, is the theory of sets, so that's all it talks about. It doesn't have the language, by itself, to discuss other kinds of objects. Class theory, underpinning category theory, is a notable exception, but most of mathematics is covered by set theory.

² According to Wikipedia, formal set theory began in the 1870s with Dedekind and Cantor. The informal idea of an ordered pair must date back at least to Descartes in the 17th century, given the Cartesian plane.

³ If you define $(a, b, c)$ to be $\{\{a\}, \{a, b\}, \{a, b, c\}\}$, as you might naively expect, you run into problems!

⁴ That said, mathematicians tend to be (deliberately) uncreative with this kind of thing, so I would suspect most would suggest similar definitions, if pressed.

⁵ Note: you could interpret $A^2$ similarly ($2 = \{0, 1\}$) as a set of functions. However, this does not obviate the need to define an ordered pair, as functions are sets of ordered pairs!

Answer 2

That depends. If you are using $A^2$ and $A^3$ to indicate the set of functions from $2=\{0,1\}\to A$, and $3=\{0,1,2\}\to A$, respectively, then the answer is a resounding "no."

If instead we define $A^n$ to be the product of $n$ copies of $A$ for each natural number $n$, the answer depends on whether you take repeated products on the left or the right. As a matter of convention, you will have either $A^2=A\times A;A^{n+1}=A\times A^n$, in which case $A\times A^2=A^3$, or $A^2=A\times A;A^{n+1}=A^n\times A$, in which case $A^3=A^2\times A\ne A\times A^2$.

From what I've seen, the most commonly used convention is $(x,y,z):=((x,y),z); A^{n+1}=A^n\times A$. However, different authors may adopt different conventions. If you are working out of a textbook, and the author uses $A^n$ to denote repeated products (as opposed to the set of functions $n\to A$), I would use whichever convention (left or right) the author uses.

Note that if we do use $A^n$ to indicate functions from the set $n$ to the set $A$, then $A^1\ne A$.

Answer 3

What we do have, and what is useful, is that there is a cononical bijection $\phi : A \times A^2 \to A^3$, where $$ \phi\big((a,(b,c))\big) = \big(a,b,c\big) $$ Whether $(a,(b,c)) = \big(a,b,c\big)$ is not important; even if it happens to true, no one (except a perverse set theorist) would care.