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?

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    $\begingroup$ The direct sum contains only the almost null sequences: sequences with at most finitely many entries different from $0$. It doesn't even contain the constant sequence $(1,1,1,\ldots)$. What you want is the direct product, and the direct product is not obviously countable (in fact, it is not countable). $\endgroup$ Apr 23, 2021 at 18:05
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    $\begingroup$ Not even the binary sequences $B := \lbrace (s_n)_{n \in \mathbb{N}}: s_n \in \lbrace 0, 1 \rbrace ~\forall n\rbrace$ form a countable set. See here. $\endgroup$
    – Meowdog
    Apr 23, 2021 at 18:07
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    $\begingroup$ Now you are saying the tensor product? If you mean $\prod \mathbb{Q}$, then the assertion that it would be countable is... problematic; in fact, it is false. $\endgroup$ Apr 23, 2021 at 18:10
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    $\begingroup$ $\oplus$ (\oplus): direct sum. Almost null tuples. $\otimes$ (\otimes): tensor product (a relatively complicated structure, not what you want here). $\prod$ (\prod): direct product, the set of all tuples. The latter is what you want. $\endgroup$ Apr 23, 2021 at 18:15
  • $\begingroup$ You know that finite cross product of countable sets is countable. But there is no reason to believe an infinite cross product of countable sets is. Although $\prod_{i=1}^n \mathbb Q$ is countable, $\prod_{i=1}^\infty \mathbb Q$ is not. $\endgroup$
    – fleablood
    Apr 23, 2021 at 19:15

1 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.

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 "$\bigoplus$."

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    $\begingroup$ See also math.stackexchange.com/q/500849 $\endgroup$ Apr 23, 2021 at 18:17
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    $\begingroup$ @ArturoMagidin ... Oh dear lord. Fixed! $\endgroup$ Apr 23, 2021 at 18:49
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    $\begingroup$ I guess it could also be that one of them is empty... ;-) $\endgroup$ Apr 23, 2021 at 19:04
  • $\begingroup$ @ArturoMagidin aaaaaaAAAAAaaaargh $\endgroup$ Apr 23, 2021 at 19:04

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