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  • A countable set is one that can be put on a one-to-one correspondence with the natural numbers. This means that you can explicitly write down a table with the ordered Naturals in one column, and each member of the countable set in the other column. You literally "count" them.

  • On the other hand, an ordered set can be such that, for any two members a and b, you can find another member c that is between them. I'm not sure what the name for this property is, so I call it "dense". (If there is another term for this please clarify).

At first glance, it seems to me that both those properties are incompatible. Edit: Why I think so: If I try to enumerate them, I would always fail at the second step: $0; 1; .. $ no wait. I forgot $0.1$. So I start over: $0; 0.1; .. $ no wait, I forgot $0,01$. So I start over: $0; 0.001; .. $no wait... etc. This won't happen with the naturals: $0; 1; 2; 3; .. $, since they are not "dense".

Yet, the set of the Rationals satisfies both. It is clear that it is "dense". And this is a way to count the Rationals. I have read it and I understand it. But I still can't wrap my mind about how can a set be countable and "dense" at the same time. To me it seems contradictory. Could someone explain it to me in layman terms?

Edit 2: As the comments have shown me, my confusion was that, I stated that it was contradictory to have a set that was dense and also had "an always-increasing bijection with the naturals". But I confused that last property, with "Countable", and that was wrong. I could ask a replacement question, if that's possible: Is it correct to state that a set that is "dense", cannot have "an always-increasing bijection with the naturals"?

Juan Perez
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    Could you please explain why you intuitively feel that those two properties should be incompatible ? – G. Fougeron May 26 '21 at 12:20
  • @G.Fougeron I added a paragraph of it. – Juan Perez May 26 '21 at 12:28
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    "If I try to enumerate them, I would always fail at the second step: 0;1;.. no wait. I forgot 0.1. So..." Who says that you have to start at a value and always go increasing? We can zig-zag... like $1,\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\dots$ – JMoravitz May 26 '21 at 12:29
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    "And [this is a way to count the rationals]. I have read it and I understand it." Apparently not since you are asking this question in this way... – JMoravitz May 26 '21 at 12:30
  • That’s because the notion of “counting” may be conflicting with the notion of “order”, that is, you may be more or less thinking that you have to count the rationals “in their order”. But you don’t have to! In the case of the rationals, here’s a way to count them all: count, for each integer $n$ (for increasing $n$), all the $k/n!$ for $-(n+1)! < k< (n+1)!$. You can see that we never exhaust any particular interval, instead we come back infinitely many times, covering more and more numbers each time. – Aphelli May 26 '21 at 12:32
  • @JMoravitz I think I got it: What I find incompatible, is that a set can have an always increasing enumeration, and also be dense. But thats not what countable means, so it's ok, and my confusion is cleared. I think. – Juan Perez May 26 '21 at 12:33
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    Indeed, any montonous enumeration cannot work. – G. Fougeron May 26 '21 at 12:34
  • "A set can have an always increasing enumeration and also be dense" Assuming by "enumeration" you mean a bijection with the naturals, the rationals do not have any such always increasing enumeration... – JMoravitz May 26 '21 at 12:34
  • I have edited my question based on these comments – Juan Perez May 26 '21 at 12:39

2 Answers2

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This should feel surprising, because it is. This is one of the first hints we encounter that infinity is a somewhat hairier concept than we might have originally hoped.

The strategy you propose for enumerating the rationals is, indeed, doomed to failure. But that is not the only such strategy - as you should know from reading the proof of the countability of the rationals. I suggest "drawing out" a successful counting strategy (e.g. the one from the proof) on a number line - this should give you some intuition about how density is reached in a countable number of steps.

user3716267
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You are trying to make the bijection order preserving ,which as you have shown is not possible it is however possible to come up with a bijection which is not monotonic as mentioned in the link .

Vivaan Daga
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