# Thread: Prove a countable Cartesian product of countable sets is uncountable

1. ## Prove a countable Cartesian product of countable sets is uncountable

Hey guys, could you please check over my solution and let me know if I made a mistake somewhere?

Problem: Let $\displaystyle A_1, A_2, A_3, ...$ be countable sets. Then show that the Cartesian product $\displaystyle A = A_1 \times A_2 \times A_3 \times ...$ is uncountable. In other words, show that the set of countably infinite sets whose elements are natural numbers is uncountably large.

Approach: I know that this is solvable using a diagonalization argument. But I wanted to try something else. I attempted to surject this to real numbers.

Consider a subset of $\displaystyle A$, say $\displaystyle B$ where $\displaystyle B$ holds all such elements $\displaystyle \{a_1, a_2, a_3, ...\}$ of A where $\displaystyle a_2,a_3,a_4,... \leq 9$.

So I defined a function $\displaystyle f: B \rightarrow \mathbb{R}$ where:

For any element of B, $\displaystyle f$ sends that set to $\displaystyle \sum_{n \in \mathbb{N}} \frac{1}{10}^{n-1} a_n = a_1 + \frac{a_2}{10} + \frac{a_3}{100} + ... a_1.a_2 a_3... \in \mathbb{R}$

This function is surjective because for all real numbers $\displaystyle x \in \mathbb{R}$ whose digits are: $\displaystyle a_1. a_2 a_3 ...$, we can find an element $\displaystyle \{a_1, a_2, a_3, ...\} \in B$

The existence of this surjection implies that $\displaystyle card(B) \geq card(\mathbb{R})$. And we also know that $\displaystyle B \subseteq A$, so that $\displaystyle card(A) \geq card(B)$. By transitivity, $\displaystyle card(A) \geq card(B) \geq card(\mathbb{R})$ which means that $\displaystyle A$ must be uncountable.

2. This makes sense to me but I'm not better than you are I think so I would wait for someone else to tell you that this is correct.
(I take part of this thread so I get an email when someone answers because it's an interesting question)

3. Originally Posted by Last_Singularity
$\displaystyle \sum_{n \in \mathbb{N}} \frac{1}{10}^{n-1} a_n$
The above one looks rational numbers, which are countable.
Plz correct me if I am wrong.

4. Hello,

For any element of B, f sends that set to $\displaystyle \sum_{n \in \mathbb{N}} \frac{1}{10}^{n-1} a_n = a_1 + \frac{a_2}{10} + \frac{a_3}{100} + ... a_1.a_2 a_3... \in \mathbb{R}$
Write \frac{1}{10^{n-1}}, this will give the correct writing.

The above one looks rational numbers, which are countable.
Plz correct me if I am wrong.
It is true that rational numbers make a countable set. However, not all number having an infinity of decimals are rational.
Irrrationals also have a decimal expansion, though it never ends and never repeats.

So I guess what Last_Singularity did is correct... like vincisonfire, don't take my word for granted ><

5. Looks all good to me although you know much better than I do. What would we do without infinite cardinality!

6. ## Re: Prove a countable Cartesian product of countable sets is uncountable

Apologies for the necro, but I found this post in a Google search on the first page of results, so I figured I should address some issues of the OP.

Originally Posted by Last_Singularity
Consider a subset of $\displaystyle A$, say $\displaystyle B$ where $\displaystyle B$ holds all such elements $\displaystyle \{a_1, a_2, a_3, ...\}$ of A where $\displaystyle a_2,a_3,a_4,... \leq 9$.
Small nitpick: an element of $\displaystyle A$ isn't a set, so don't use set notation - use parentheses to indicate that it is a tuple or sequence.
Larger nitpick: it was never mentioned that the various countable sets $\displaystyle A_n$ were subsets of the integers. Countable sets needn't necessarily be comprised of natural numbers. You need to do some legwork to get to a usable set of numbers: let us define $\displaystyle S_n$ as the "substitute" set, and wherever $\displaystyle A_n$ and $\displaystyle a_n$ appears in the proof, substitute it instead by $\displaystyle S_n$ and $\displaystyle s_n$. Given that each $\displaystyle A_n$ is countable, there is a bijection from a subset of the natural numbers. If this subset is not infinite, then it has cardinality $\displaystyle c_n$ - in this case, let $\displaystyle S_n$ be the set of natural numbers strictly less than $\displaystyle c_n$. Otherwise, let $\displaystyle S_n = \mathbb{N}$.

For any element of B, $\displaystyle f$ sends that set
Again, elements of $\displaystyle A$ (and thus $\displaystyle B$) are not sets, they are sequences or tuples. If it were a set, it would have a cardinality of at most 11, given how you chose $\displaystyle B$.

This function is surjective because for all real numbers $\displaystyle x \in \mathbb{R}$ whose digits are: $\displaystyle a_1. a_2 a_3 ...$, we can find an element $\displaystyle \{a_1, a_2, a_3, ...\} \in B$
No it isn't. To be surjective, you require further assumptions. As a counter example, let each $\displaystyle A_n$ be the countable set $\displaystyle \{0, 1\}$. The infinite countable product $\displaystyle \{0, 1\}^\mathbb{N}$ is uncountable, but your "proof" does not show this. Your proof actually gives a weaker result.

To complete you proof, you need the following assumptions. From the collection of countable sets $\displaystyle A_n$, there must be at least infinitely many sets with at least 10 elements - this lets us have infinitely many digits from 0 to 9. We would then "discard" any of the sets which do not have at least 10 elements - we can get them back later by taking the cartesian product of the resultant uncountably infinite set with the discarded sets. Unless any of these discarded sets is the empty set (yet another assumption which isn't mentioned), the product would also be uncountably infinite. Additionally, there must be at least one set with infinitely many elements - this would correspond to your $\displaystyle A_1$ in the above proof. Finally, obviously, you must assume that $\displaystyle \mathbb{R}$ is uncountably infinite. This final assumption is a pretty big sticking point - how would you go about showing that $\displaystyle \mathbb{R}$ is uncountable? If you manage it without some variant of the diagonalization argument, then that would be quite interesting, but I haven't yet seen a proof that isn't just diagonalization with different words. So your proof boils down to a weaker result that inevitably stems from diagonalization anyway.

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