Question: Constructing a Subset of the Real Numbers with Cardinality Aleph 1

[lavinia]

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]

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

 

[lavinia]

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]

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]

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]

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]

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.

 

[Office_Shredder]

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]

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.

 

[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]

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]

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]

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

 

[CRGreathouse]

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

See post #2. No such explicit example can be constructed.

 

[lavinia]

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.

 

[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]

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.

 

[yossell]

What's the claim, CR? I'm not following you. That you can't construct an example? Or that you can `construct' an example, but it gives you no insight? Perhaps Lavinia could be the judge of whether she finds insightful or not.

Whether or not the continuum hypothesis is true, one can give examples of aleph_1 subsets of the real numbers. Just take a mapping from the reals between 0 and 1 onto the ordinals; then consider that subset of these reals which, by this function, are mapped onto aleph_1.

Of course, I agree that this is an uninteresting example, for many reasons. One reason is that this subset may be nothing more than the interval (01) again. Another reason is that the construction goes through whether or not continuum hypothesis holds, and so may not have been what Lavinia wanted when it was explicitly assumed that CH failed. I agree that, at least for me, there's no insight here - at least for me.

Lavinia seemed to make it clear that he/she wasn't asking for an explicit constructible example - whatever that means (do you think the reals are constructible?). But (IMO) Lavinia's right that it's just not true that a description of the relevant set somehow contradicts the undecidability of the CH.

 

[lavinia]

What's the claim, CR? I'm not following you. That you can't construct an example? Or that you can `construct' an example, but it gives you no insight? Perhaps Lavinia could be the judge of whether she finds insightful or not.

Whether or not the continuum hypothesis is true, one can give examples of aleph_1 subsets of the real numbers. Just take a mapping from the reals between 0 and 1 onto the ordinals; then consider that subset of these reals which, by this function, are mapped onto aleph_1.

Of course, I agree that this is an uninteresting example, for many reasons. One reason is that this subset may be nothing more than the interval (01) again. Another reason is that the construction goes through whether or not continuum hypothesis holds, and so may not have been what Lavinia wanted when it was explicitly assumed that CH failed. I agree that, at least for me, there's no insight here - at least for me.

Lavinia seemed to make it clear that he/she wasn't asking for an explicit constructible example - whatever that means (do you think the reals are constructible?). But (IMO) Lavinia's right that it's just not true that a description of the relevant set somehow contradicts the undecidability of the CH.

Thanks.

There is an example Rudin's Real and Complex Analysis where he assumes that the Continuum Hypothesis is true - which means that there is a well ordering of the reals that maps them bijectively onto the first uncountable ordinal. Given this well ordering he defines a set by the rule, {(x,y) such that x<y in the well ordering}. The exercise is to show that this set is not Lebesque measurable.

 

[micromass]

Let's presume that the continuum hypothesis is false. Then there exists a set X of cardinality \aleph_1 in \mathbb{R}. So it is indeed natural to ask whether X would be Lebesguemeasurable and stuff.

Even with assuming the negation of CH, there is still not much we can say about X. Luckily we can accept other axioms. If we, for example, adopt Martins axiom. Then X will be Lebesgue measurable and it will have measure zero. It will also then be of first category.

 

[yossell]

Let's presume that the continuum hypothesis is false. Then there exists a set X of cardinality \aleph_1 in \mathbb{R}.

as I keep saying, there exists a set of cardinality \aleph_1 in \mathbb{R} whether the continuum hypothesis is true or false. <br /> <br /> Ok - maybe I&#039;m wrong - but could someone tell me why.

 

[CRGreathouse]

What's the claim, CR? I'm not following you. That you can't construct an example? Or that you can `construct' an example, but it gives you no insight?

That you can 'construct' but cannot construct. In the same worthless way that you can demonstrate, say, a nonprinciple ultrafilter using AC (*pop* one exists! but we don't know anything about it!), we can worthlessly demonstrate an aleph_1 subset of R with not-CH (*pop* one exists! but we know nothing about it!).

 

[CRGreathouse]

there exists a set of cardinality \aleph_1 in \mathbb{R} whether the continuum hypothesis is true or false

<br /> <br /> Correct. I don&#039;t know why you keep bringing this up, though. I think it was evident to everyone participating on this thread before you mentioned it.

 

[yossell]

That you can 'construct' but cannot construct.

I see. Got it. That's great. It's totally clear now. That's very helpful.

Correct. I don't know why you keep bringing this up, though. I think it was evident to everyone participating on this thread before you mentioned it

Thanks. I brought it up because the argument seemed to be `suppose CH is false. Then there is a subset of the reals which is of cardinality aleph-1'. But the supposition does no real work in getting the conclusion. So I was puzzled by the role of the supposition in people's thought.

 

[CRGreathouse]

Thanks. I brought it up because the argument seemed to be `suppose CH is false. Then there is a subset of the reals which is of cardinality aleph-1'. But the supposition does no real work in getting the conclusion. So I was puzzled by the role of the supposition in people's thought.

In post #25, for example, the assumption is needed for the "know nothing about it" claim. With CH, we can explicitly construct sets of cardinality aleph_1; without, we cannot.

 

[yossell]

In post #25, for example, the assumption is needed for the "know nothing about it" claim. With CH, we can explicitly construct sets of cardinality aleph_1; without, we cannot.

I'm really perplexed. Could you help me by explaining your notion of explicit constructibility?

(a) Are you thinking of Godel's axiom of constructibility? Here, I at least see the link between constructibility and CH, as CH is implied by this axiom.

(b) Do you think sets can only be explicitly presented without use of the axiom of choice? I see that when sets are defined (constructed?) without this axiom, we almost always have a far greater amount of information about the set defined (constructed?) without. But I wouldn't say that we have no information about sets defined using the axiom of choice.

(c) Do you have a constructivism in mind as found in http://en.wikipedia.org/wiki/Constructivism_(mathematics ) ?

(d) Why, if CH holds, do you think we can `construct' aleph_1. Do you think the power set operation (on an aleph_zero sized set) is a `constructible' operation (which, under CH, would give us an aleph_1 sized set)? Why does this operation count as `constructible' while taking all the well-orderings of an aleph-zero sized set (which always gives us an aleph-one sized set whether or not CH holds) is not a `constructible' operation?

TL;DR:

I don't really understand your link between your notion of `construction' and your notion of `know nothing about it'. I've some guesses, but none of them really explain it.

 

[lavinia]

I think that this thread has lost its seam. The original question that I asked is not being discussed.