Flawed proof that real numbers are countable

Question

I know there is something wrong with this argument, I am just not sure where.

Consider constructing the real numbers are equivalence classes of cauchy sequences of rationals. So in some sense the real numbers are a subset of equivalence classes of cauchy sequences of rationals so lets just consider those. In fact lets just consider all sequences of rationals.

A sequence of rationals has countably infinite many terms. Each term has countably infinite possibilities (any rational numbers). So the number of sequences is really just the same amount as

$$\bigotimes_{i=1}^\infty\mathbb{Q} $$ which would be countable.

Where is the flaw here?

Answer 1

While the union of countably many countable sets is countable and the product of finitely many countable sets is countable, the product of countably many countable sets is not countable (unless all but finitely many of those sets are singletons, or at least one of those sets is empty).

So you are right that $$\mathbb{R}\equiv\prod_{i\in\mathbb{N}}\mathbb{Q}$$ ("$\mathbb{Q}^\mathbb{N}$" would also be appropriate notation here, but the symbol "$\otimes$" refers to something quite different). However, your error is that this does not imply that $\mathbb{R}$ is countable.

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As a quick aside:

Let's say we have a set $X$ with some distinguished element $x_0$. We can consider the subset of $X^\mathbb{N}$ consisting of all sequences of elements of $X$ which are "mostly $x_0$," that is, such that all but finitely many elements of the sequence are $x_0$. This describes a countable set, and is basically analogous to how the countable set of reals with terminating (= eventually $0$) decimal expansions sits inside the uncountable set $\mathbb{R}$.

This sort of operation comes up a lot in abstract algebra; the relevant term is "direct sum" (as opposed to product), and this is denoted by "$\bigoplus$."