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I understand that you can create a bijection from $\mathbb{N}$ to $\mathbb{Z}$ such that to every natural number you can assign an integer (e.g. this one). Therefore, one knows that both sets have the same cardinal.

The problem I encounter is that even when you can "make pairs" and not leave any element from either set, it's also true that one set has more elements than the other. Namely:

  1. If we consider $0\in\mathbb{N}$, then $\mathbb{Z}-\mathbb{N}=\{-1,-2,-3,...\}$.
  2. Else: $\mathbb{Z}-\mathbb{N}=\{0,-1,-2,-3,...\}$.

So can we really say that both have the same size but one has more elements than the other? What is the real meaning? Is size in this case defined not as the number of elements in both sets but in their tendency? That is, if they tend to have the same number of elements, not if they have the same exact quantity.

Antoniou
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    The "real meaning" of size here (when saying they have the same size) is "cardinality". It is true. The "real meaning" of "one has mere elements than the other" is "its cardinality is greater". It is false. The "tendency" of a set, or "they tend to have the same number of elements", means nothing. – Anne Bauval Jun 20 '23 at 15:35
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    Maybe this could help you? Why do the rationals, integers and naturals all have the same cardinality? – Bruno B Jun 20 '23 at 15:36
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    In contrast to the situation with finite sets, for infinite sets there are several inequivalent notions of "size", and mixing them together inevitably leads to confusion. In the sense of cardinality $\mathbb N$ and $\mathbb Z$ are the same size: there exists a one-to-one correspondence (a.k.a a bijection) between $\mathbb N$ and $\mathbb Z$. – Joe Jun 20 '23 at 15:37
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    In the sense of inclusion, we might say that $\mathbb Z$ is strictly larger than $\mathbb N$, since $\mathbb N$ is a proper subset of $\mathbb Z$. The drawback of using inclusion to measure the "size" of sets is that there are many sets $A$ and $B$ for which $A$ is not a subset of $B$, and $B$ is not a subset of $A$, and so they cannot be compared. By contrast, you can always ask whether $A$ and $B$ can be put into one-to-one correspondence. – Joe Jun 20 '23 at 15:37
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    It is characteristic of an infinite set that it can have a subset of the same "size". So, if you want to avoid this weirdness then you need to stick with finite sets. Search for Hilbert's Hotel. This might help you understand. – badjohn Jun 20 '23 at 15:53
  • Some discussion of Hilbert's Hotel here: https://math.stackexchange.com/questions/3557700/can-there-be-an-irrational-numbers-hotel/3557752#3557752 – badjohn Jun 20 '23 at 16:26
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    So can we really say that both have the same size but one has more elements than the other? They are of the same cardinality but one is a strict subset of the other. What is the real meaning? Depends what you call "real" but I would say look up the definitions of "same cardinality" and "subset" and don't use vague terms like "size" and "more elements". –  Jun 20 '23 at 16:28
  • They are exactly the same size. Leave zero as is and relabel the positive integers $1,3,5,..$ and relabel the negative integers to be $2,4,6,8$. Then the proper subset relationship is still maintained because ${0,1,3,\dots}$ gets mapped to $\mathbb N\subset\mathbb Z$. So yes, they have the same elements with different names. – John Douma Jun 20 '23 at 17:37
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    Seems weird for beginners with infinite sets , but it is well known that a proper subset (or superset) can have the same cardinality. Even the set of algebraic numbers is still countable , hence has the same cardinality as the set of natural numbers. – Peter Jun 20 '23 at 17:42

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