Apparent contradiction of the diagonal argument - where am I going wrong?

Question

Please show me where I am going wrong...

Let $G_{n}$ be the $n$-dimensional, infinite, unit-square-grid , i.e. the grid whose points lie in $\mathbb{N}^n$ separated by a distance of $1$.

Assign every dimension a unique $n^{th}$ prime $p_n$

And assign every point on every axis, the value of the power of $n$ equal to that point's distance from the origin. So on the $2-$axis we have $(1,2,4,8,\ldots)$ and on the $3-$axis we have $(1,3,9,27,\ldots)$

Now assign every point within the n-grid the value of the product of its co-ordinates drawn off all axes.

As we take $n\to\infty$ the primes are enumerated by the axes and every positive, whole number greater than $1$ is enumerated by the points within the grid, according to the fundamental theorem of arithmetic, plus $1$ which lies at the origin.

The cardinality of the count of the axes is countably infinite.

The cardinality of the number of points on every axis is countably infinite.

The cardinality of the points on this infinite-dimensional grid of infinite length, is $N^N$ is it not, which is uncountable?

But these points are enumerated by the integers - a contradiction of $\lvert \mathbb{N}\rvert<\lvert \mathcal P(\mathbb{N})\rvert$.

Where am I going wrong?

The only possibility I can see is that the primes could be uncountably infinite. But more likely I'm making a simple error.

Answer 1

This is essentially the same mistake as thinking that you can map the reals to the naturals by "reversing the digits" of a real to get a natural: most of the "naturals" so created will be infinite.

In this case, note that in the infinite-dimensional grid, most points will have infinitely many nonzero coordinates. The "natural" assigned to such a point will be a product of infinitely many numbers $>1$ - and this isn't actually a natural number. E.g. what number do you assign to $(1, 1, 1, 1, ...)$?

There's a "phase transition" that happens when you move from the finite-dimensional case to the infinite-dimensional case. You're aware of part of it - the number of points in the grid jumps from countable to uncountable - but it also has other ramifications, e.g. the inability to appropriately label points by individual numbers.

Incidentally, there is a better notion of infinite-dimensional grid that makes the counting work! Namely, let $G_\infty$ be the set of all points in the infinite-dimensional grid with only finitely many nonzero coordinates. This is analogous to the direct sum of a family of groups, as opposed to their direct product. And note that in a precise sense, $G_\infty$ really is the "union" of the $n$-dimensional grids as $n\rightarrow\infty$.

Now we can label each axis with a prime, and thus assign to each point in $G_\infty$ a unique natural number. But of course $G_\infty$ is countable - the relationship between $G_\infty$ and the full infinite-dimensional grid is exactly the relationship between the set of all finite binary strings and the set of all infinite binary strings.

Answer 2

This works for any finite number of dimensions. But when you take it to an infinite number of dimensions, consider the point $(1,1,1,\dots)$. The number you're trying to assign to this point is $2\cdot3\cdot5\cdot\cdots$, which doesn't work since it diverges to infinity. This method does work to enumerate every point with only a finite number of non-zero entries, and indeed there are only a countable number of such points.