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I am having some trouble with an intuitive understanding of how we can say two sets equal in cardinality iff there is a bijection between them.

In particular, a bijection exists between $\mathbb{Z}$ and $\mathbb{N}$; however, $\mathbb{N} \subseteq \mathbb{Z}$ but not $\mathbb{Z} \subseteq \mathbb{N}$, yet we claim $|\mathbb{N}| = |\mathbb{Z}|$. (1)

From a basic POV: $\mathbb{N} \subseteq \mathbb{Z} \implies |\mathbb{N}| \le |\mathbb{Z}|$, but I've been unable to arrive at a contradiction; since in the other case, the premise is false.

I'm having trouble what basic intuitive, primitive permits us to have the above situation, denoted (1) be possible -- outside of being purely based on definition.

FD_bfa
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Anonymous Anonymous
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  • Using inclusion to compare the size of two sets is not a good idea. Which set would have more elements, the set of even integers or the set of odd integers? – jjagmath Jan 02 '25 at 02:22
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    You are quite correct, though, about the fact that $$A\subseteq B\implies\lvert A\rvert\leq\lvert B\rvert$$ for any two sets $A$ and $B.$ – Cameron Buie Jan 02 '25 at 02:48
  • What are ZNF axioms? – Martin Brandenburg Jan 02 '25 at 06:53
  • @jjagmath if you mean even and odd integers on the whole $\mathbb{Z}$, they would have equal number of elements. You can compare sizes of much more complex sequences: https://arxiv.org/abs/2411.00296 – Anixx Jun 03 '25 at 15:59
  • I know they have the same size, that's not the point of my comment. – jjagmath Jun 03 '25 at 21:48

1 Answers1

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The issue you are facing is that infinite sets do not behave in the same way as finite sets. Your question asks for intuition, but this is simply the way things are defined.

For finite sets $A \subset B \implies |A| < |B|$. For infinite sets, this is not necessarily true. The way to resolve the problem you are facing is to realise that when talking about two infinite sets having the same cardinality, the subset argument changes to the following: $A \subset B \implies |A| \color{red}{ \leq} |B|$.

In the comments under @jjagmath's answer you attempt to use a subset argument to show that $|\mathbb{N}| < |\mathbb{Z}|$, but the above shows that we can only get as far as $|\mathbb{N}| \color{red}{ \leq}|\mathbb{Z}|$, which you know is true.

The two different implications for finite and infinite sets using subset arguments are both consequences of the fact that two sets are defined to have the same cardinality if and only if there is a bijection between them. This is the definition. The subset arguments above are just consequences of this definition.

FD_bfa
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  • Thank you for the answer. That was clear. Could you point me towards a resource where I could see how the difference in the subset arguments you describe arises from the definition? – Anonymous Anonymous Jan 03 '25 at 01:37
  • The best way to see this is through examples (many have been asked on this site). For example, this answer shows that the cardinality of even numbers is equal to the cardinality of all natural numbers (https://math.stackexchange.com/questions/341605/as-many-even-numbers-as-natural-numbers). This is despite the fact that one is clearly a subset of the other. @AnonymousAnonymous – FD_bfa Jan 03 '25 at 03:06
  • For an example not involving the natural numbers, note that the interval $(0,a)$ for any $a>0$ has equal cardinality to $\mathbb{R}$ (see, e.g., https://math.stackexchange.com/questions/1896320/cardinality-of-0-1-and-mathbbr) – FD_bfa Jan 03 '25 at 03:09
  • Thank you. I was more curious if there was a general proof of the fact. Something like the two statements you made, one for finite and one for infinite. – Anonymous Anonymous Jan 03 '25 at 03:18
  • For any sets $A$ and $B$, if $A \subset B$, then by definition $|A| \leq |B|$ (this is true for finite and infinite sets) @AnonymousAnonymous – FD_bfa Jan 03 '25 at 03:22
  • For finite sets, we can strengthen this argument since $B \setminus A \neq \emptyset \implies B \neq A \implies |A| < |B|$. I can elaborate on any of those steps if you need me to, but this is essentially a complete proof. – FD_bfa Jan 03 '25 at 03:22
  • For that last bit, wouldn't that also apply to $\mathbb{N} = A$ and $\mathbb{Z} = B$? – Anonymous Anonymous Jan 03 '25 at 03:53
  • The proof doesn't work for infinite sets because it relies on the fact that $B$ must have at least one more element than $A$. In other words, we know $|A| + 1 = |B|$ implies that $|A| < |B|$. But (informally), $\infty + 1 = \infty$ and so the argument breaks down if you try this for $\mathbb{N}$ and $\mathbb{Z}$ @AnonymousAnonymous – FD_bfa Jan 03 '25 at 09:10