笛卡尔积的序同构问题：$N\times R$与$R\times R$字典序是否同构？

Great question! Let's unpack this by looking at the key structural differences between the two ordered sets—these differences will show they can't be order-isomorphic.

First, let's recap the lexicographic order definition for both sets:
For any pairs $(a,b), (c,d)$:

• $(a,b) < (c,d)$ if either $a < c$, or $a = c$ and $b < d$.

Now, let's break down the critical distinctions:

1. The "block" structure of each set #

Both sets can be partitioned into maximal subsets (let's call them "blocks") where every pair of elements in a block is bounded above and below by another element in the same block. For:

• $\mathbb{N} \times \mathbb{R}$: Each block is ${(n, r) \mid r \in \mathbb{R}}$ for some fixed $n \in \mathbb{N}$. These blocks are ordered exactly like $\mathbb{N}$: block $n$ comes strictly before block $n+1$, and there are no blocks between block $n$ and block $n+1$. In short, the blocks form a discrete, countable ordered set (isomorphic to $\mathbb{N}$).
• $\mathbb{R} \times \mathbb{R}$: Each block is ${(r, s) \mid s \in \mathbb{R}}$ for some fixed $r \in \mathbb{R}$. These blocks are ordered exactly like $\mathbb{R}$: for any two blocks corresponding to $r_1 < r_2$, there's always a block for some $r_3$ where $r_1 < r_3 < r_2$. Here, the blocks form a dense, uncountable ordered set (isomorphic to $\mathbb{R}$).

2. Why this breaks order-isomorphism #

An order-isomorphism is a bijective function that preserves the order relation. This means it must map blocks in $\mathbb{N} \times \mathbb{R}$ to blocks in $\mathbb{R} \times \mathbb{R}$, while preserving the order of the blocks themselves.

But $\mathbb{N}$ (the order type of $\mathbb{N} \times \mathbb{R}$'s blocks) is not order-isomorphic to $\mathbb{R}$ (the order type of $\mathbb{R} \times \mathbb{R}$'s blocks):

• $\mathbb{N}$ has discrete elements (each element has an immediate successor), while $\mathbb{R}$ is dense (no element has an immediate successor).
• $\mathbb{N}$ is countable, $\mathbb{R}$ is uncountable.

Since the block structures can't be mapped to each other via an order-preserving bijection, the entire sets can't be order-isomorphic either.

Another way to see it: Cofinality of initial segments #

Take any element in $\mathbb{N} \times \mathbb{R}$ like $(k, 0)$. The set of elements less than $(k,0)$ is the union of $k-1$ blocks (each isomorphic to $\mathbb{R}$) plus a subset of the $k$-th block. That's a countable union of sets each isomorphic to $\mathbb{R}$ (or a subset of $\mathbb{R}$).

In contrast, take any element in $\mathbb{R} \times \mathbb{R}$ like $(s, 0)$. The set of elements less than $(s,0)$ is the union of uncountably many blocks (one for every $r < s$) plus a subset of the $s$-th block. An order-isomorphism would have to map a countable union of these structures to an uncountable one, which is impossible—order-isomorphisms preserve the cardinality of such decompositions.

内容的提问来源于stack exchange，提问作者m3mir