Just a quick question: Is it correct to say that the set of rational numbers cannot be a subset of the set of natural numbers? Certainly, we know these two sets have the same cardinality and there exist a bijection between them. But yet, we cannot say the set of rational numbers is contained in (or is a subset of) the set of natural numbers right? Since, after all, we certainly can find a rational number that does not belong in the set of natural numbers. I think I just am mixing up some simple notions of subset or equal sets vs isomorphic sets under some bijective mapping but it's confusing me.
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2 Answers
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It seems to me that your confusion is between $A\subset B$ and $\lvert A\rvert\leqslant\lvert B\rvert$ (that is, the cardinal of $A$ is smaller than or equal to the cardinal of $B$). In turns out that $A\subset B\implies\lvert A\rvert\leqslant\lvert B\rvert$, but this is not an equivalence.
In your specific situation, $\mathbb Q$ and $\mathbb N$ have the same cardinal, but, as you know, $\mathbb{Q}\not\subset\mathbb N$.
answered Sep 12, 2018 at 18:02
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$\begingroup$ By "this is not an equivalence", you mean the converse of that implication is not true in general yea. Think it is clearer to me now thanks! $\endgroup$ Commented Sep 12, 2018 at 18:07
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$\begingroup$ @user82479 Yes, that is what I mean. $\endgroup$ Commented Sep 12, 2018 at 18:08
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$$\frac12\in\mathbb Q\setminus\mathbb N.$$
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$\begingroup$ Certainly, this is easy to see but I am just having difficulty resolving how we can say both sets are isomorphic and yet the rationals are not a subset of the naturals? So the first involves mapping whereas the second involves just elements, thus it doesn't make sense to compare these two questions? $\endgroup$ Commented Sep 12, 2018 at 18:01
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$\begingroup$ @user82479: a bijection does not require the the two sets be "compatible", i.e. to share elements. $\endgroup$– user65203Commented Sep 12, 2018 at 18:03