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I've read that such is the case but I've thought up of a bijection which makes me think otherwise. My apologies if this has been asked before...

Let $\mathbb{W} = \mathbb{N} \cup \{0\}$. Let $P$ be the set of primes. Denote by $p_{k}$ the $k$th prime number.

Let $\{W_{p_{n}}\}_{n=1}^{\infty}$ be a sequence such that $W_{p_{i}} = \mathbb{W}$ for all $i \in \mathbb{N}$.

Then $S := W_{p_{1}} \times W_{p_{2}} \times \cdots = \mathbb{W} \times \mathbb{W} \times \cdots$ is a Cartesian product of countably infinite copies of $\mathbb{W}$, since the set of primes $P$ is countably infinite.

Now, consider $f : S \rightarrow \mathbb{N}$ such that $f(w_{1},w_{2},\dots) = p_{1}^{w_{1}}p_{2}^{w_{2}}\cdots = 2^{w_{1}}3^{w_{2}}\cdots$

We know that the every natural number has a unique prime factorization, so that $f$ is a bijection, from where it follows that $S$ is numerically equivalent to $\mathbb{N}$.

Where is the fault in this proof?

Andrés E. Caicedo
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David Rodriguez
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    The element $(1,1,1,\dots) \in S$ but the function $f$ maps it to $\prod_{p} p$ which is not a natural number. So $f$ is not "well-defined". –  Sep 22 '16 at 18:50
  • "We know that the every natural number has a unique prime factorization, so that f is a bijection" But the range of f is not $\mathbb N$. It is a set for which $\mathbb N$ is a subset. for (w1,w2,w3....) infinite then $f(w1,w2,.....) \not \in \mathbb N$. – fleablood Sep 22 '16 at 19:07
  • "Now, consider f:S→N such that f(w1,w2,…)=pw11pw22⋯=2w13w2⋯" That's not $f: S -> N$ that is $f:S->2^N$. – fleablood Sep 22 '16 at 19:09

2 Answers2

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What if the "number" $f(w_1, w_2, . . .)$ is infinite? That is, if infinitely many of the $w_i$s are nonzero.

Indeed, the set is uncountable: can you find a surjection onto $2^\mathbb{N}$?

What you have shown is that the part of that infinite product which consists of sequences all but finitely many of whose terms are zero, is countable.

Noah Schweber
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"Where is the fault in this proof?"

The infinite product of an infinite number of primes is not a natural number.

That's it.

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This is subtle but it's important: The infinite collection of finite things is not the same as the collection of infinite things. This is why the infinite collection of finite decimal expansions (which is a subset of the rationals) is not the same thing as the collection of infinite decimal expansions (which is the reals). And why the set of natural numbers is itself infinite even though every single on of the infinite number of natural numbers is actually finite.

So the set of cartesion products up to any of the infinitely possible finite degrees can map to a finite prime product, the set of cartesion products to an infinite degree will map to an infinite prime product which is not a natural number. In fact, it's a nonsensical concept.

fleablood
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