# The Continuum Hypothesis

#### lavinia

Gold Member
Suppose we assume that the Continuum Hypothesis is false. Then there must be a subset of the real numbers that has the cardinality of Aleph 1. What is an example of such a subset?

#### Office_Shredder

Staff Emeritus
Gold Member
If you could construct such a set then the continuum hypothesis wouldn't be undecidable

#### lavinia

Gold Member
If you could construct such a set then the continuum hypothesis wouldn't be undecidable
Can you explain this more?

It would seem that if say the Continuum were Aleph2, then there would be a subset of cardinality Aleph1 just as there is a subset of cardinality Aleph0.

#### Office_Shredder

Staff Emeritus
Gold Member
The continuum hypothesis is neither false nor true under the ZF axioms. If you could construct an uncountable set with cardinality smaller than c, then the continuum hypothesis would obviously be false.

You might think that if it's impossible to construct such a set, they must not exist. But with only finitely many characters and statements of finite length, there are only countably many things we can say. So you can't construct every subset of the real numbers. So a subset whose cardinality is smaller than c might exist but we're unable to actually construct it

#### lavinia

Gold Member
The continuum hypothesis is neither false nor true under the ZF axioms. If you could construct an uncountable set with cardinality smaller than c, then the continuum hypothesis would obviously be false.

You might think that if it's impossible to construct such a set, they must not exist. But with only finitely many characters and statements of finite length, there are only countably many things we can say. So you can't construct every subset of the real numbers. So a subset whose cardinality is smaller than c might exist but we're unable to actually construct it
I am a little confused here. If the unit interval had a cardinality greater than the first uncountable ordinal then it would still have to have a proper subset that has the cardinality of the first uncountable ordinal just as it has countably infinite subsets.

I thought uncountable just meant greater cardinality than that of the integers not the same cardinality as the continuum.

I wonder though if such an uncountable subset could have positive Lebesque measure.

#### Office_Shredder

Staff Emeritus
Gold Member
That doesn't mean you can explicitly construct it. Similar to how the axiom of choice is used to state the existence of things that otherwise would not be constructible

#### lavinia

Gold Member
That doesn't mean you can explicitly construct it. Similar to how the axiom of choice is used to state the existence of things that otherwise would not be constructible
You said that if you could construct such a set the continuum hypothesis would obviously be false. Why is that?

can you tell me anything about it? E.g. would it be Lebesque measurable?

Even if it is not constructible you may still be able to describe its properties. For instance if you assume that the Continuum Hypothesis is true, then you can describe a non-measurable subset of the cartesian plane even though you may not be able to construct a mapping the continuum onto the first uncountable ordinal.

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#### Office_Shredder

Staff Emeritus
Gold Member
If you can construct an uncountable set with cardinality smaller than c, then there exists a set with cardinality smaller than c. The continuum hypothesis says no such set exists, so it must be that the continuum hypothesis is false.

I don't really know a whole lot about the subject. There was a talk I went to a couple years ago where someone described how there's a whole class of theorems which are of the form "If A is a subset of the real numbers with property X, then A is either has the first uncountable cardinality or has the continuum as its cardinality" and of course half the people interpret this as meaning the continuum should be the second uncountable cardinality, and half the people interpret it as meaning the continuum should be the first

#### yossell

If you can construct an uncountable set with cardinality smaller than c, then there exists a set with cardinality smaller than c. The continuum hypothesis says no such set exists, so it must be that the continuum hypothesis is false.
To be fair to the OP, that's not literally what was said in the first post. All that was asked was for an example of a subset of the real numbers that had cardinality Aleph 1. But a subset S of the real numbers, even a proper subset, doesn't necessarily have a lower cardinality than the real numbers. There would only be a problem with the unprovability of the continuum hypothesis if you could also show that subset S had a cardinality less than the continuum, but greater than aleph zero.

Having said that, whether or not you assume the continuum hypothesis, there is a subset of the real numbers that has cardinality Aleph_1. (We just can't prove that there is such a subset has a cardinality less that the continuum). So it's not clear exactly what the original poster had in mind with the question.

#### lavinia

Gold Member
To be fair to the OP, that's not literally what was said in the first post. All that was asked was for an example of a subset of the real numbers that had cardinality Aleph 1. But a subset S of the real numbers, even a proper subset, doesn't necessarily have a lower cardinality than the real numbers. There would only be a problem with the unprovability of the continuum hypothesis if you could also show that subset S had a cardinality less than the continuum, but greater than aleph zero.

Having said that, whether or not you assume the continuum hypothesis, there is a subset of the real numbers that has cardinality Aleph_1. (We just can't prove that there is such a subset has a cardinality less that the continuum). So it's not clear exactly what the original poster had in mind with the question.
My question was simple - I thought. If we assume that the Continuum Hypothesis is false then there must be a proper subset of the line of the plane that has cardinality Aleph 1. Constructibility had nothing to do with the question. All I wanted to know was what such a subset might look like and maybe even an example. By example I did not mean construct per se but more some defining rule.

I wonder if the properties of this set depend strongly on the particular well ordering of the reals.

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#### yossell

If we assume that the Continuum Hypothesis is false then there must be a proper subset of the line of the plane that has cardinality Aleph 1.
But, strictly speaking, whether or not the continuum hypothesis is false, there is a proper subset of the real line that has cardinality aleph 1. For even under the assumption that the continuum hypothesis is true, and the real line has cardinality aleph 1, there are still proper subsets of the real line that have the same cardinality as the real line (for instance, the points between 0 and 1/2), and thus (under the assumption) have cardinality aleph 1.

#### chronon

From a countable set of reals you can generate a new real by diagonalization, and you can repeat this as often as you like. You can then mirror the generation of countable ordinals to get a 1-1 correspondence between the set of countable ordinals (i.e aleph 1) and a subset of the real numbers (although I'm not sure that this correspondence counts as being properly defined)

#### George Jones

Staff Emeritus
Gold Member
lavinia, what Office_Shredder has assumed, is that instead of
Suppose we assume that the Continuum Hypothesis is false. Then there must be a subset of the real numbers that has the cardinality of Aleph 1. What is an example of such a subset?
you meant something like
Suppose we assume that the Continuum Hypothesis is false. Then there must be a subset of the real numbers that has the cardinality of Aleph 1, and such that there is no bijection between the subset and the set of real numbers. What is an example of such a subset?
Is this what you meant? yossell has pointed out that these are quite different things. This is similar to the following.

The integers are a proper subset of the reals such that there is no bijection between the integers and the reals.

#### chronon

If the continuum hypothesis is false and S is a subset of the reals of cardinality aleph 1 then you can deduce immediately that there is not a bijection between S and the reals.

#### George Jones

Staff Emeritus
Gold Member
To be fair to the OP, that's not literally what was said in the first post. All that was asked was for an example of a subset of the real numbers that had cardinality Aleph 1. But a subset S of the real numbers, even a proper subset, doesn't necessarily have a lower cardinality than the real numbers.
No, in the original post, lavinia asked for an cardinality Aleph 1 subset of the reals together with the condition that
we assume that the Continuum Hypothesis is false
and, as chronon points out in post #14,
If the continuum hypothesis is false and S is a subset of the reals of cardinality aleph 1 then you can deduce immediately that there is not a bijection between S and the reals.
Thus,
these are quite different things.
is not correct.

#### lavinia

Gold Member
These responses are all interesting but the question still stands unanswered.

#### CRGreathouse

Homework Helper
These responses are all interesting but the question still stands unanswered.
See post #2. No such explicit example can be constructed.

#### lavinia

Gold Member
See post #2. No such explicit example can be constructed.
The properties of such a set could potentially be described without an explicit construction. One might appeal to the Well Ordering Principle to obtain a mapping of the reals onto one of the uncountable ordinals whose cardinality is possible for the continuum.

From there one might be able to say something about a subset of cardinality Aleph1 e.g. maybe that it must have Lebesque measure zero.

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#### disregardthat

The properties of such a set could potentially be described without an explicit construction. One might appeal to the Well Ordering Principle to obtain a mapping of the reals onto one of the uncountable ordinals whose cardinality is possible for the continuum.

From there one might be able to say something about a subset of cardinality Aleph1 e.g. maybe that it must have Lebesque measure zero.
The continuum hypothesis is undecidable even in ZFC, so you could not construct (nor "construct" using the well ordering principle) such a subset. It is undecidable; it has been shown that it cannot be proved nor disproved in ZFC. The bridge cannot be built from the axioms to the statement.

#### CRGreathouse

Homework Helper
The continuum hypothesis is undecidable even in ZFC, so you could not construct (nor "construct" using the well ordering principle) such a subset. It is undecidable; it has been shown that it cannot be proved nor disproved in ZFC. The bridge cannot be built from the axioms to the statement.
Right. You could 'construct' it from ¬CH in the same sense that you can 'construct' a well-ordering of the reals with AC, but it's not going to give you any insight.

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