Math Help - cardinality of the Cantor set only uncountable or necessarily that of the continuum?

1. cardinality of the Cantor set only uncountable or necessarily that of the continuum?

In the Wikipedia account of the Cantor tertiary set, it is remarked that the cardinality of the set is uncountable. Can this can be strengthened to it needing to be as large as the continuum? (This is not necessarily the same thing if we assume that the continuum hypothesis is false, so that uncountable could also be aleph-1, below the continuity of the continuum.) It would seem by the proof given there that it would be, but I am not certain. If one started out with a subset of the interval with only aleph-one points in it, would the construction go through?
Thanks.

2. The cantor set $T$ is homeomorphic to the product space:

$\prod \left\{{A_i:i\in \mathbb{N}}\right\},\quad A_i=\left\{{0,2}\right\}$

where in $A_i$ we consider the discrete topology, so:

$\#(T)=2^{\aleph_0}=c$

Fernando Revilla

3. Thanks. In other words, yes to my first question and no to the second one.

By the way, there seems to be something grammatically wrong with your quote from Sylvester. "is according to superficially" doesn't make sense. Is there a typo there?

By the way, there seems to be something grammatically wrong with your quote from Sylvester. "is according to superficially" doesn't make sense. Is there a typo there?

It is an exact quote written in not contemporary English. Look for example here

Fernando Revilla

5. Nomadreid - look at Fernando's post carefully. He is telling you that the cantor set has the same cardinality as the continuum. That is, it is the same size as the reals. It is not $\omega_1$ unless the continuum hypothesis holds.

6. Originally Posted by DrSteve
Nomadreid - look at Fernando's post carefully. He is telling you that the cantor set has the same cardinality as the continuum. That is, it is the same size as the reals. It is not $\omega_1$ unless the continuum hypothesis holds.

Is $\omega_1$ a new way to denote $\aleph_1$? I really don't know.

Tonio

7. Originally Posted by tonio
Is $\omega_1$ a new way to denote $\aleph_1$? I really don't know.

Tonio
$\omega_1$ is the notation usually used for the ordinal, and $\aleph_1$ is used for the cardinal. Of course these two numbers are equal (because $\omega_1$ is a cardinal and therefore it is equal to its cardinality), so they can be used interchangeably. I think that I might be one of the few people that uses the ordinal notation - I developed this habit in graduate school simply because writing the hebrew letter was unnatural for me and ruined my flow.

And it's not a new notation - I believe this notation was used since the ordinals were introduced.

8. Originally Posted by DrSteve
$\omega_1$ is the notation usually used for the ordinal, and $\aleph_1$ is used for the cardinal. Of course these two numbers are equal (because $\omega_1$ is a cardinal and therefore it is equal to its cardinality), so they can be used interchangeably. I think that I might be one of the few people that uses the ordinal notation - I developed this habit in graduate school simply because writing the hebrew letter was unnatural for me and ruined my flow.

And it's not a new notation - I believe this notation was used since the ordinals were introduced.

I know the ordinal notation for the naturals in the usual order, but never saw it used in the way you did.

I can't see how an ordinal can be the same as a cardinal as they both apply on different things.

For exmple, the set $\{2,....,1\}$ is cleary of cardinality $\aleph_0$ , yet its ordinal is $\omega+1$ , and

it can't be that $\omega+1=\aleph_0$ since then also $\omega=\aleph_0$ by taking the

naturals in the usual order...

Tonio

9. Every well-ordered set is isomorphic to a unique ordinal. The example you gave is a well-ordered set isomorphic to the ordinal $\omega +1$. The ordinal $\omega + 1$ is not a cardinal. It's cardinality is $\omega$ or as you probably prefer to call it $\aleph_0$. So $\omega +1 \ne \omega$ because these are nonisomorphic well-ordered sets. But the cardinality of $\omega + 1$ is $\omega$ because there is a bijection between $\omega +1$ and $\omega$.

From the set theoretic point of view every cardinal is an ordinal. A cardinal is just an ordinal that is not equinumerous with any smaller ordinal (A and B are equinumerous if there is a bijection between them).

Given an arbitrary set, assuming the axiom of choice, this set has a unique cardinality. The cardinality of the set is a cardinal (thus it is also an ordinal). Note that to "see" cardinality you just need a bijection, whereas to see which ordinal you are isomorphic to requires an order-preserving bijection.

As one more example (to try to help clarify my use of notation), every countable ordinal (and in fact every countable set) has cardinality $\aleph_0$, but only the ordinal $\omega$ is actually equal to $\aleph_0$. thus it is mathematically correct to use $\omega$ in place of $\aleph_0$. The difference between them is only the way you think about them. People tend to think about $\aleph_0$ as the size of the natural numbers, and as $\omega$ as the ordering of the natural numbers, but in fact they are the same (at least when using the standard set-theoretical definitions).

Note: I realize that you probably understand most of this already - I'm just trying to give you enough information to justify my use of notation.

10. DrSteve, my reply to him indicated that I understood him, so I am not sure what the correction is about. Let me reiterate: I asked two questions:
(1) "....the cardinality of the set is uncountable. Can this can be strengthened to it needing to be as large as the continuum?"
(2) " If one started out with a subset of the interval with only aleph-one points in it, would the construction go through?" ( I meant, of course, a dense subset of the unit interval.)
I answered " yes to my first question and no to the second one". This is what you wrote, so there is no disagreement.

11. Aha - I completely missed the second question. Thanks for the clarification.

12. Originally Posted by DrSteve
Every well-ordered set is isomorphic to a unique ordinal. The example you gave is a well-ordered set isomorphic to the ordinal $\omega +1$. The ordinal $\omega + 1$ is not a cardinal. It's cardinality is $\omega$ or as you probably prefer to call it $\aleph_0$. So $\omega +1 \ne \omega$ because these are nonisomorphic well-ordered sets. But the cardinality of $\omega + 1$ is $\omega$ because there is a bijection between $\omega +1$ and $\omega$.

From the set theoretic point of view every cardinal is an ordinal. A cardinal is just an ordinal that is not equinumerous with any smaller ordinal (A and B are equinumerous if there is a bijection between them).

Given an arbitrary set, assuming the axiom of choice, this set has a unique cardinality. The cardinality of the set is a cardinal (thus it is also an ordinal). Note that to "see" cardinality you just need a bijection, whereas to see which ordinal you are isomorphic to requires an order-preserving bijection.

As one more example (to try to help clarify my use of notation), every countable ordinal (and in fact every countable set) has cardinality $\aleph_0$, but only the ordinal $\omega$ is actually equal to $\aleph_0$. thus it is mathematically correct to use $\omega$ in place of $\aleph_0$. The difference between them is only the way you think about them. People tend to think about $\aleph_0$ as the size of the natural numbers, and as $\omega$ as the ordering of the natural numbers, but in fact they are the same (at least when using the standard set-theoretical definitions).

Note: I realize that you probably understand most of this already - I'm just trying to give you enough information to justify my use of notation.

Yes, I knew that but your explanation helped quite a bit. I still am not convinced

of that equivalence in using both term interchangeably since I'm used to think of

ordinals as intrinsically attached to order, whereas cardinals are not, but of course

this may be just my own ignorance.

Tonio

13. Originally Posted by tonio
Yes, I knew that but your explanation helped quite a bit. I still am not convinced

of that equivalence in using both term interchangeably since I'm used to think of

ordinals as intrinsically attached to order, whereas cardinals are not, but of course

this may be just my own ignorance.

Tonio

The most accepted definition of cardinality in current times seems to be that the cardinality of a set is the least ordinal to which that set is isomorphic. This definition depends on the axiom of choice, since without choice some sets can't be well-ordered. I always follow this definition.

I believe that some would strip away the ordering in the definition of cardinal (as you seem to do). If you take this stance, then you would be correct in saying that my notation is incorrect.

I believe that if choice fails, a definition of cardinality could still be made using the axiom of foundation, but I have not studied this myself (I always assume choice).

14. Originally Posted by tonio
I can't see how an ordinal can be the same as a cardinal as they both apply on different things.
With the most common set theoretical definitions, every cardinal is an ordinal. A cardinal is an ordinal that is not equinumerous with any of its members.

15. Originally Posted by tonio
I still am not convinced of that equivalence in using both term interchangeably
They're not used interchangably. Every cardinal is an ordinal, but not every ordinal is a cardinal.

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