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This question arose while attempting to prove the proposition in the provided link.

Because the cardinality of a set formed by the Cartesian product of finitely many sets of natural numbers is always countable, we cannot simply equate a set $X$ with $\mathbb{N}$ just because a bijection can be established between $X$ and $\mathbb{N}$.

This is because, in certain aspects, $\mathbb{N}^2$ and $\mathbb{N}$ still differ. For instance, in the problem mentioned above, I believe there are far more ways to sum over $\mathbb{N}^2$ (i.e., summing over an infinite table with infinitely many rows and columns) than to sum over $\mathbb{N}$ (which only has a single infinite row of numbers). For example, when summing over an infinite table, I could first add up the countable numbers in the first row before considering the subsequent rows. Therefore, I want to find a property that can distinguish $\mathbb{N}^2$ from $\mathbb{N}$ (and even $\mathbb{N}^n$).

I came across online discussions about distinguishing $\mathbb{R}^k$ of different dimensions, which involved topological isomorphism, and I wonder if there are corresponding conclusions for $\mathbb{N}^2$ and $\mathbb{N}$ (and even $\mathbb{N}^n$).

……Cantor's result does not mean that the concept of dimension is merely an illusion that should be abandoned. On the contrary, his discovery reveals the existence of a dividing line beyond which purely set-theoretic concepts no longer apply and should be replaced by other concepts with different properties. In 1910, Brouwer proved that the field of topology is capable of distinguishing between different dimensions.……

Supplement

The motivation behind this question

the reason I raised this question from issues I encountered regarding the order of summation when dealing with nonnegative functions $a_x$ defined on a countable set $X$(i.e.$\sum_{x\in X}^{}a_x $). For instance, when summing an infinite two-dimensional array, we can first sum each row and then sum the column totals. However, without introducing an extension of rearrangement definitions via infinite ordinals, this approach cannot be described as a common rearrangement. Just as we cannot claim that summing all odd-indexed terms first and then the even-indexed terms in a one-dimensional sequence constitutes a rearrangement of the original sequence. Yet, I wish to formulate such operations under a unified framework. Therefore, I believe it is necessary to distinguish between $\mathbb{N}$ and its Cartesian products.

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    $\begingroup$ The default topology on $\mathbb{N}^n$ for $n \in \mathbb{N}$ is the discrete topology, so topological dimension of any kind is not going to be able to tell them apart. That is, if you accept that a singleton is zero-dimensional, and the disjoint sum of zero-dimensional spaces is zero-dimensional. $\endgroup$ Commented Nov 11 at 7:03
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    $\begingroup$ In short, from the topological POV, all $\mathbb N^n$ are identical and have no such thing as dimension, while $\mathbb R^n$ are all different. $\endgroup$ Commented Nov 11 at 7:08
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    $\begingroup$ What, exactly, do you mean by distinguish? Surely you see that the sets are different (in fact disjoint). So I assume you mean to ask whether they are (non-)isomorphic in some sense, but then you have to specify which properties you care for. E.g. they are isomorphic as topological spaces, but not as monoids. $\endgroup$ Commented Nov 11 at 7:38
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    $\begingroup$ It is not really clear what you are asking. As abelian semigroups, $\mathbb N^n$ are all different and characterized by their rank. That's certainly a way of distinguishing among them. $\endgroup$ Commented Nov 11 at 7:40
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    $\begingroup$ Subsets don't offer any new perspective. All sets of similar cardinality have the same subsets. Representations of anything are not a property of that thing at all (you number them with tuples, someone else numbers them with white rabbits). Pretty much everything in math are sets with some structure, and the possible structures are numerous and diverse. You tried the structure of topology, and that failed to tell apart $\mathbb N$ from $\mathbb N^2$. Think of something else. Partially ordered sets, by any chance? Linear spaces? Then there are groups, fields and a ton of other things. $\endgroup$ Commented Nov 11 at 7:48

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As the commenters say, topology is probably not what you're looking for. From many perspectives, the most natural topology to put on $\mathbb{N}$ is the discrete topology, and from this perspective $\mathbb{N}$ is homeomorphic (read: indistinguishable as a topological space) from $\mathbb{N}^n$ for any $n$. The picture to have in mind is that up to "stretching and squishing" (read: topology) there's no difference between infinitely many dots in a line and infinitely many dots in a grid (read: $\mathbb{N}$ and $\mathbb{N}^2$).

However we can use the order relation $\leq$ to get our hands on the phenomenon you're interested in! Let's write $x \lessdot y$ to mean that $x < y$ and there's nothing in between. So for instance in $(\mathbb{N},\leq)$ we have $2 \lessdot 3$ but $2 \not \lessdot 4$. Precisely, we say

$$(x \lessdot y) \iff (x < y \land \lnot \exists z . x < z < y).$$

But what's happening in $\mathbb{N}^2$ with the usual product order where $(x_1,x_2) \leq (y_1,y_2)$ if and only if $x_1 \leq x_2$ and $y_1 \leq y_2$? Now we see that, for instance, $(2,3) \lessdot (3,3)$ and $(2,3) \lessdot (2,4)$ -- do you see why? The order is keeping track of the relative positions of the dots, and for $\mathbb{N}^2$ this $\lessdot$ relation is sensitive to the fact that there's "two directions" to move in. This is making precise your observation about the difference between a "table with infinitely many rows and columns" and a "single infinite row". After all, we might think of $(p+1,q)$ as incrementing the row and $(p,q+1)$ as incrementing the column, and we're able to get our hands on this purely order theoretically by using $\lessdot$.

More generally can you show that in $\mathbb{N}^n$ with its pointwise $\leq$ order, you can recover $n$ from the $\lessdot$ relation?

I hope this helps ^_^

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  • $\begingroup$ Thank you,It's very help and that is a answer I want,^_^ $\endgroup$ Commented Nov 11 at 8:11

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