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Let's say that we divide the region $(0,1)$ into $N$ bins of width $1/N$. Of course, it makes sense to take the limit $1/N \rightarrow 0$ in this configuration, because that's simply how we define an integral (the limit of the sum of evaluations of a function [times $1/N$] at the beginning of each bin, as $1/N \rightarrow 0$ is simply the integral of that function).

What I'm wondering is, does it make sense to take the limit of the cardinality of the set of bins? That is, let $S$ be the set of bins. Does it make sense to look at the limit $\lim\limits_{1/N \rightarrow 0} |S|$?

If so, would that limit be $\aleph_0$ or $\aleph_1$? I ask because there's an intuitive reason to believe either one. If you take the limit as the width becomes zero, then you could say that there's a bin at every conceivable point in $(0,1)$, which means that the set of bins should be bijective with the set $(0,1)$ which would lead me to think that the limit is $\aleph_1$. However, throughout the entire limit, the number of bins is countable, and it's hard to imagine how a countable set can discontinuously jump to an uncountable set right as the limit is reached, which would lead me to think that the limit is $\aleph_0$.

Do does the limit make sense, and if so, what is that limit?

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  • $\begingroup$ The cardinality of $\mathbb R$ is $2^{\aleph_0}$, whether that is equal to $\aleph_1$ is an open question known as the continuum hypothesis. $\endgroup$ – Henrik supports the community Apr 16 '15 at 16:36
  • $\begingroup$ @Henrik It's still known that $2^{\aleph_0} > \aleph_0$, just not whether it's the smallest number that's greater than $\aleph_0$. I just sort of assumed the continuum hypothesis, but if I take that assumption away and simply replace every instance of $\aleph_1$ with $2^{\aleph_0}$ my question is still valid. Is the limit of the cardinality of the set of bins as bin width approaches 0 countably infinite or uncountably infinite? $\endgroup$ – Bridgeburners Apr 16 '15 at 16:45
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    $\begingroup$ The limit is neither. Cardinals are not real numbers. $\endgroup$ – Asaf Karagila Apr 16 '15 at 16:48
  • $\begingroup$ It's hard to define the limit of sets. And then its totally unclear wheter limit commutes e.g. with taking cardinality $\endgroup$ – Hagen von Eitzen Apr 16 '15 at 17:08
  • $\begingroup$ Yes, your question is valid - I just wanted to make sure you knew that the cardinality might not be $\aleph_1$. I think you specify the construction of the bins more precisely for lengths that are not $1/n$, but I don't know if that will actually help. $\endgroup$ – Henrik supports the community Apr 16 '15 at 19:49
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If you must insist on an answer it would be $\aleph_0$, but the question itself is wrong.

The reason the question is wrong is that cardinals are not real numbers. Not even finite cardinals. Yes, I am well-aware that $1+1=2$ both in the real numbers and in cardinal arithmetic. But when you leave finitary operations of finite cardinals, namely when you go to the limit, then you can easily see why the two system are different.

But even if you insist that since finite cardinals are in fact real numbers, then the answer of that limit is $\infty$ rather than $\aleph_0$ or $\aleph_1$. Because the $\aleph$ number have absolutely nothing to do with real analysis and calculus. And indeed the limit of $\lim_{n\to\infty} n=\infty$.

That been said, the cardinals themselves carry their own order topology which means we can in fact talk about limits. Since this sequence is monotone, this is the same as asking what is $\sup\{n\mid n<\aleph_0\}$, and indeed the answer would be $\aleph_0$.

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  • $\begingroup$ This does reflect my limited knowledge of set theory. But to be fair to myself, I didn't know whether the question was proper which is why I made sure to ask "does the limit make sense and if so...." I made it clear that I didn't know whether the limit was well defined, so I made that my initially contingent question. Nevertheless, I appreciate your answer. It does perfectly answer my question. $\endgroup$ – Bridgeburners Apr 16 '15 at 17:50
  • $\begingroup$ You're welcome. $\endgroup$ – Asaf Karagila Apr 16 '15 at 17:52

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