# Are real numbers Countable?

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TeethWhitener
Gold Member
I trust that you understand that 'uncountable' is a 'yes-or-no' condition, and that therefore nothing can possibly be "even more uncountable" than something else that is 'uncountable' is; however, I think that it's worth pointing out that even the infinitesimal is exactly as 'uncountably infinite' in its fractional expansion as the entirety of ##\mathbb{R}## (the reals) in its is.
I think @Svein is referring to the fact that the power set of ##\mathbb{R}## has a greater cardinality than ##\mathbb{R}##.

I think @Svein is referring to the fact that the power set of ##\mathbb{R}## has a greater cardinality than ##\mathbb{R}##.
I think that you're right about that; however, I trust that he understands that the greater cardinality does not increase its uncountability, which of course cannot increase beyond 'uncountable' any more than a switched-on light switch can be more 'on' than 'on'.

Is this correct?

The cardinality of an interval in R is proportional to its length, and as it is uncountable, the cardinality of some arbitrarily small nonzero interval in R is greater than all of N

Is this correct?

The cardinality of an interval in R is proportional to its length, and as it is uncountable, the cardinality of some arbitrarily small nonzero interval in R is greater than all of N
The cardinality of any non-empty interval of non-zero length in ##\mathbb{R}## is not proportional to its length; it's the same as that of the whole of ##\mathbb{R}## (and is therefore greater than that of ##\mathbb{N}##). Although the length of 1 inch on a 12-inch ruler is obviously less than that of the 12 inches of such a ruler, the cardinality of the unit interval ##[0,1]## is the same as that of the interval ##[0,12]##. The number of possible divisions is uncountably infinite in both cases. A discussion of matters fundamentally related to this can be found at https://plato.stanford.edu/entries/continuum-hypothesis/

dextercioby and BWV
Svein
I trust that you understand that 'uncountable' is a 'yes-or-no' condition
Some jokes fall more than flat.

PeroK
FactChecker
Gold Member
Some jokes fall more than flat.
I think we should give some "artistic license" to the terminology used and recognize that @Svein has made an important point.

Dale
Some jokes fall more than flat.
When you check your pockets on leaving the house ##-## let's see, keys, wallet, phone, TI-83 (graphing calculator), ok, good to go ##-## is a too-flat ##-## hmm ##-## more than flat ##-## maybe a B♭ (B flat) clarinet that's more than too flat maybe an A♮ (A natural) clarion?

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dextercioby
Homework Helper
I think @Svein is referring to the fact that the power set of ##\mathbb{R}## has a greater cardinality than ##\mathbb{R}##.

Is the cardinality of the powerset of ##\mathbb{R}## (or ##\mathbb{R}^n##, with ##n## finite) aleph_2?
If not, what is aleph_2?

stevendaryl
Staff Emeritus
Is this correct?

The cardinality of an interval in R is proportional to its length, and as it is uncountable, the cardinality of some arbitrarily small nonzero interval in R is greater than all of N

No, that's not correct. The cardinality of an interval is not affected by its length. An interval of length 1 has the same cardinality as the full set of real numbers.

Two sets have the same cardinality if you can map one set to the other in a one-to-one fashion. The mapping:

##x \mapsto log(\frac{x}{1-x})##

maps any ##x## in the range ##(0,1)## to a real number in the range ##(-\infty, +\infty)##, showing that those two ranges have the same cardinality.

Dale, sysprog, FactChecker and 1 other person
Gazing up into the darkness I saw myself as a creature driven and derided by vanity; and my eyes burned with anguish and anger.

No, that's not correct. The cardinality of an interval is not affected by its length. An interval of length 1 has the same cardinality as the full set of real numbers.

Two sets have the same cardinality if you can map one set to the other in a one-to-one fashion. The mapping:

##x \mapsto log(\frac{x}{1-x})##

maps any ##x## in the range ##(0,1)## to a real number in the range ##(-\infty, +\infty)##, showing that those two ranges have the same cardinality.
So do (0,1) and (0,12) in R have the same cardinality because a function exists to map them (multiply/divide by 12) or do, by definition, all uncountable sets have the same cardinality?

by the above, do (0,1) and (0,12) in the rationals have the same cardinality?

stevendaryl
Staff Emeritus
So do (0,1) and (0,12) in R have the same cardinality because a function exists to map them (multiply/divide by 12) or do, by definition, all uncountable sets have the same cardinality?

No, there are uncountable sets with cardinality larger than the cardinality of all real numbers.

The question of whether there are uncountable sets with cardinality smaller than the cardinality of all real numbers is an unsolved problem. And unsolvable. Given the usual tools of mathematics, there is no way to prove or disprove the claim that the set of all real numbers is the smallest uncountable cardinality.

But certainly all sets of reals that are going to come up in ordinary mathematics are in one of three categories:

1. Finite sets.
2. Countably infinite sets.
3. Uncountably infinite sets.

by the above, do (0,1) and (0,12) in the rationals have the same cardinality?

Yes, every infinite set of rationals has the same cardinality.

sysprog and BWV
Infrared
Gold Member
Is the cardinality of the powerset of ##\mathbb{R}## (or ##\mathbb{R}^n##, with ##n## finite) aleph_2?
If not, what is aleph_2?

Allowing for ##n\geq 2## doesn't change anything since ##\mathbb{R}## and ##\mathbb{R}^n## have the same cardinality.

This question is an example of the generalized continuum hypothesis, which is known to be independent from ZFC (https://en.wikipedia.org/wiki/Continuum_hypothesis#The_generalized_continuum_hypothesis).

sysprog
TeethWhitener
Gold Member
Is the cardinality of the powerset of R (or Rn, with n finite) aleph_2?
If not, what is aleph_2?
This depends on whether your system accepts or rejects the continuum hypothesis. The cardinality of the natural numbers is ##\aleph_0## and the cardinality of the real numbers is ##2^{\aleph_0}##, but the continuum hypothesis (the assertion that ##2^{\aleph_0}=\aleph_1##) is independent of the axioms of ZFC.

Dali, Hornbein and sysprog
FactChecker
Gold Member
do, by definition, all uncountable sets have the same cardinality?
This part is not right. There are higher cardinal numbers. There is a whole sequence of them of increasing size.

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BWV
Cantor proved that the power set of ##\mathbb{R}## has cardinality greater than that of ##\mathbb{R}##. The continuum hypothesis is that nothing has cardinality in between those. Please if you find such a thing, or can prove that there is no such thing, let us know.

WWGD, Dale and FactChecker
TeethWhitener
Gold Member
do, by definition, all uncountable sets have the same cardinality?
Just to pile on, Cantor's theorem is that, for any set ##A##, given its powerset ##\mathcal{P}(A)##--defined as the set of all subsets of ##A##--the cardinality of the powerset will always be strictly greater than the cardinality of the set: ##|\mathcal{P}(A)|>|A|##

Klystron, sysprog and BWV
WWGD
Gold Member

1 - 1
2 - 2
3 - 3
...
10 - 10
11 - 11
...
123-123
...

Assume that this list contains all natural numbers. Now add 1 to the ones place of the first number.(1+1=2), add 1 to the tens place of the second number(0+1=1), add 1 to the hundreds place of the third number(0+1=1), and so on. (If you do it third times, you get 112.)

The number you get, if you concatenate all changed digits is a number that was not in your list, i.e. whichever list you choose, there is a number that is not listed. Hence the natural numbers are uncountable.
What's the image of ##\pi##? Or any other Irrational?
Sorry @emptyboat , I thought I was replying to the OP.

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WWGD
Gold Member
Cantor proved that the power set of ##\mathbb{R}## has cardinality greater than that of ##\mathbb{R}##. The continuum hypothesis is that nothing has cardinality in between those. Please if you find such a thing, or can prove that there is no such thing, let us know.
Good point. It's true for every set, finite or infinite.

So do (0,1) and (0,12) in R have the same cardinality because a function exists to map them (multiply/divide by 12) or do, by definition, all uncountable sets have the same cardinality?

by the above, do (0,1) and (0,12) in the rationals have the same cardinality?
There is also measure theory which is probably more along the lines of your original intuition.

https://en.wikipedia.org/wiki/Lebesgue_measure#:~:text=Any closed interval [a, b,b and has measure zero.&text=Moreover, every Borel set is Lebesgue-measurable.

BWV
stevendaryl
Staff Emeritus
There is also measure theory which is probably more along the lines of your original intuition.

Except that standard measure theory gives 0 as the measure for any collection of rational numbers. So it doesn't support the intuition that there are twice as many rationals in (0,2) as in (0,1).

mathwonk
Homework Helper
2020 Award
but it is consistent with it since twice zero is zero. ???

stevendaryl
WWGD
Gold Member
So do (0,1) and (0,12) in R have the same cardinality because a function exists to map them (multiply/divide by 12) or do, by definition, all uncountable sets have the same cardinality?

by the above, do (0,1) and (0,12) in the rationals have the same cardinality?
Yes, two sets have the same cardinality iff there is a bijection between them.

BWV
WWGD said:
BWV said:
So do (0,1) and (0,12) in R have the same cardinality because a function exists to map them (multiply/divide by 12) or do, by definition, all uncountable sets have the same cardinality?

by the above, do (0,1) and (0,12) in the rationals have the same cardinality?
Yes, two sets have the same cardinality iff there is a bijection between them.
That is of course true; however, it is clearly not true that "all uncountable sets have the same cardinality", ##-## e.g. ##|2^\mathbb{R}|>|\mathbb{R}|## (i.e., the cardinality of the power set of ##\mathbb{R}## is strictly greater than that of ##\mathbb{R}##), and both ##2^\mathbb{R}## and ##\mathbb{R}## are uncountable sets.

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WWGD