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Sometimes I talk about math to my friends, and some (with an engineering background) aren't used to the idea that in math, we have to define stuff. For example, they may not be used to the idea that $\sum_{n=0}^{\infty} a_n$ doesn't have an a priori meaning.

To get the point across, I would like to have an example to give these friends about infinite sets. Under the standard (and very natural) definition, we say that two sets have the same cardinality if they can be put into a bijection to one another. Is there any other reasonable sounding definition of cardinality which still says that (for example) $\{a, b, c\}$ has $3$ elements, but says that (for example) $\mathbb{N}$ and $\mathbb{Q}$ have different cardinalities?

Noble Mushtak
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Ovi
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  • "Reasonable" is a subjective thing. What sounds reasonable to you might not to me. – Robert Israel Jan 06 '19 at 15:48
  • @RobertIsrael Of course; perhaps I should've tagged it as a soft question. But I hope that approximately everyone will approximately agree on what is approximately reasonable. – Ovi Jan 06 '19 at 15:50
  • I feel like the concept you are trying to demonstrate is much easier to do with series rather than the cardinality of sets. For example, $\sum_{n=0}^\infty (-1)^n$ is not a convergent sum if you define convergence as the limit of the partial sums, but does converge under the Cesaro sum, which is the limit as $n\to\infty$ of the arithmetic mean of the first $n$ partial sums. This shows how, although both of these definitions of convergence are somewhat reasonable, they lead to very different results. – Noble Mushtak Jan 06 '19 at 16:04
  • @NobleMushtak Yes. But I'm also curious for myself if there are any other alternative definitions for cardinality out there. – Ovi Jan 06 '19 at 16:12
  • Maybe you need to get your hands on the contents of Cantor's wastepaper basket before he came up with the proper version . . . Were other definitions attempted? (Maybe only by Cantor since he was so far ahead of his time.) – timtfj Jan 06 '19 at 18:13
  • Without the axiom of choice the idea of the cardinal number for a set can change (I know of no alternative definitions for cardinality itself). If we define the cardinal number of a set to be ordinal number of the shortest wellordering of the set, then there may be sets that have no cardinal number. If instead you use Scott's trick you can define a cardinal number of a set even if it can't be wellordered. – Malice Vidrine Jan 06 '19 at 19:25
  • Whatever your alternate definition would be a definition of, it isn't cardinality. – Robert Israel Jan 06 '19 at 19:53
  • @RobertIsrael Yes sorry, I didn't know how to say it without making the title too wordy. What I really mean is an alternative definition of when two sets have the same cardinality; I hope even this is right. – Ovi Jan 06 '19 at 21:35
  • If your friends are engineers, they should be well accustomed to the idea that $\infty+1=\infty+\infty=\infty$. – Asaf Karagila Jan 06 '19 at 22:08
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    Also, https://math.stackexchange.com/a/40318/622 https://math.stackexchange.com/a/125428/622 https://math.stackexchange.com/q/596028/622 https://math.stackexchange.com/q/168258/622 https://math.stackexchange.com/q/242057/622 https://math.stackexchange.com/q/1589887/622 and there are probably quite a few more of these roaming around here. – Asaf Karagila Jan 06 '19 at 22:10
  • @AsafKaragila Wow thanks gonna check those out! – Ovi Jan 06 '19 at 22:11
  • "If you wish to discourse with me, you must define your terminology"--- Voltaire..... Regarding $\sum_n a_n, $ we can embed $\Bbb R$ into a larger ordered field $\Bbb R^$ in which the basic interactions between $<$ and $+,\times$ still hold, but in which $\sum_{j\in \Bbb N}2^{-j}$ has no meaning because $ {x\in \Bbb R^ :x>0 \land \forall n\in \Bbb N,(1-x>\sum_{j=1}^n2^{-j})}$ is not empty and has no largest member.... $\sum_{j\in \Bbb N}2^{-j}$ has a meaning in $\Bbb R$ because of how we define $\Bbb R.$ – DanielWainfleet Jan 14 '19 at 01:24

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