Infinity

**wonderboy1953** · January 9th 2010, 06:20 AM

I understand how the set of transcendental numbers is greater than the set of algebraic numbers (Aleph II vs Aleph I). Can someone explain why the set of functions (Aleph III) is greater than the set of transcendental numbers?

**CaptainBlack** · January 9th 2010, 07:34 AM

Originally Posted by wonderboy1953

I understand how the set of transcendental numbers is greater than the set of algebraic numbers (Aleph II vs Aleph I). Can someone explain why the set of functions (Aleph III) is greater than the set of transcendental numbers?

You are using $\aleph$ in an unconventional manner $\aleph_0$ usually denote the cardinality of the naturals and $\mathfrak{c}$ that of the continuum (the reals). The continuum hypothesis is that $\mathfrak{c}=\aleph_1$ (that is that there is no set with cadinality between that of naturals and that of the reals).

The algebraic numbers are in fact countable and so have cardinality $\aleph_0$ .
The trancendentals have cardinality $\mathfrak{c}$ as they are those reals which are not algebraic.

The set of all functions on the reals is equivalent to the power set of the reals and we know that the cardinality of the power set of a set is strictly larger than that of the set itself.

CB

**Drexel28** · January 9th 2010, 11:13 AM

Originally Posted by CaptainBlack

You are using $\aleph$ in an unconventional manner $\aleph_0$ usually denote the cardinality of the naturals and $\mathfrak{c}$ that of the continuum (the reals). The continuum hypothesis is that $\mathfrak{c}=\aleph_1$ (that is that there is no set with cadinality between that of naturals and that of the reals).

The algebraic numbers are in fact countable and so have cardinality $\aleph_0$ .
The trancendentals have cardinality $\mathfrak{c}$ as they are those reals which are not algebraic.

The set of all functions on the reals is equivalent to the power set of the reals and we know that the cardinality of the power set of a set is strictly larger than that of the set itself.

CB

Agreed. http://www.mathhelpforum.com/math-he...rdinality.html

**Drexel28** · January 9th 2010, 11:44 AM

Originally Posted by wonderboy1953

I understand how the set of transcendental numbers is greater than the set of algebraic numbers (Aleph II vs Aleph I). Can someone explain why the set of functions (Aleph III) is greater than the set of transcendental numbers?

Also, how do you know that the cardinality of the algebraic numbers is greater than that of the transcendental? I mean, it's definitely true. But, I have never seen you post or say anything about cardinal numbers. Do you have proof?

**CaptainBlack** · January 9th 2010, 12:05 PM

Originally Posted by Drexel28

Also, how do you know that the cardinality of the algebraic numbers is greater than that of the transcendental? I mean, it's definitely true. But, I have never seen you post or say anything about cardinal numbers. Do you have proof?

err.. em transcendentals > algebraic ?

**Drexel28** · January 9th 2010, 12:13 PM

Originally Posted by CaptainBlack

err.. em transcendentals > algebraic ?

Oops, typo.

I meant to say that the cardinality of the transcendentals is greater than that of the algebraic! And that the proof of this isn't all together trivial!

**wonderboy1953** · January 12th 2010, 08:11 AM

I now recall that G. Gamow wrote a book titled "One, Two, Three...Infinity" where he stated that the set of functions is greater than the set of transcendentals.

Does anyone have that book and know what he actually stated?

**Drexel28** · January 12th 2010, 11:37 AM

Originally Posted by wonderboy1953

I now recall that G. Gamow wrote a book titled "One, Two, Three...Infinity" where he stated that the set of functions is greater than the set of transcendentals.

Does anyone have that book and know what he actually stated?

No, but as we stated. That is correct. We know that the transendentals have a cardinal number greater than $\aleph_0$ and less than or equal to $\mathfrak{c}$ . But, the set of all functions defined on the reals has cardinal number $2^{\mathfrak{c}}$ so that if $\text{card }\mathbb{T}=\mathfrak{m}$ then $\aleph_0<\mathfrak{m}\leqslant\mathfrak{c}<2^{\mat hfrak{c}}$ and what you said ofllows.

**wonderboy1953** · January 19th 2010, 12:49 PM

"Also, how do you know that the cardinality of the algebraic numbers is greater than that of the transcendental? I mean, it's definitely true. But, I have never seen you post or say anything about cardinal numbers. Do you have proof?"

The proof that the cardinality of the reals is greater than that of the algebraic numbers is a well-known proof given by Cantor. The proof itself is referred to as the diagonalization proof whereby (expressed in decimal form), you go down diagonally through the list of algebraic numbers to come up with a new number which isn't part of the list (which you already know is referred to as C for the reals).

I'm sure that someone proved that the cardinality of C and the reals match up (maybe Cantor). What I had originally wanted to know is how someone knew that the cardinality of the functions is greater than the cardinality of the real numbers. My answer comes from a book written by Tobias Dantzig titled "number: the language of science", copyrighted 2/2007. On Page 233 it lists as an example of an aggregate with a power greater than C the functional manifold,"i.e. the totality of all correspondences which can be established between two continua....The corresponding cardinal number is denoted by f."

It appears that all of the functions themselves are part of the foregoing definition. The book goes on to say that the functional "space" is derived from the continuum (real numbers) through the diagonalization process that led from the algebraic numbers to the real numbers. And this diagonalizational process can indefinitely produce greater aggregates.

**Drexel28** · January 19th 2010, 02:17 PM

Originally Posted by wonderboy1953

"Also, how do you know that the cardinality of the algebraic numbers is greater than that of the transcendental? I mean, it's definitely true. But, I have never seen you post or say anything about cardinal numbers. Do you have proof?"

The proof that the cardinality of the reals is greater than that of the algebraic numbers is a well-known proof given by Cantor. The proof itself is referred to as the diagonalization proof whereby (expressed in decimal form), you go down diagonally through the list of algebraic numbers to come up with a new number which isn't part of the list (which you already know is referred to as C for the reals).

I'm sure that someone proved that the cardinality of C and the reals match up (maybe Cantor). What I had originally wanted to know is how someone knew that the cardinality of the functions is greater than the cardinality of the real numbers. My answer comes from a book written by Tobias Dantzig titled "number: the language of science", copyrighted 2/2007. On Page 233 it lists as an example of an aggregate with a power greater than C the functional manifold,"i.e. the totality of all correspondences which can be established between two continua....The corresponding cardinal number is denoted by f."

It appears that all of the functions themselves are part of the foregoing definition. The book goes on to say that the functional "space" is derived from the continuum (real numbers) through the diagonalization process that led from the algebraic numbers to the real numbers. And this diagonalizational process can indefinitely produce greater aggregates.

I gave you a link to where I proved that the cardinality of $\mathbb{R}^{\mathbb{R}}$ is $2^{\mathfrak{c}}$ . The uncountability of the transcendentals is more instructively proven, in my humble opinion, by showing that the algebraics are countable (many easy ways to do this) and showing by contradiction that $\mathbb{R}-\mathbb{A}$ is uncountable.

$\mathbb{R}^n$ can also be shown to have cardinality $\mathbb{c}$ . This shown by first noting that $[0,1]^{n}\simeq [0,1]$ and then that $\mathbb{R}^n\simeq[0,1]^n$ .

Also, it can be shown that the set of all continuous mappings from the reals into the reals has cardinality $\mathbb{c}$ .

Also,...I digress.

If you would like proofs or just justifications for any of these claims. Let me know.

**CaptainBlack** · January 19th 2010, 02:22 PM

Originally Posted by wonderboy1953

"Also, how do you know that the cardinality of the algebraic numbers is greater than that of the transcendental? I mean, it's definitely true. But, I have never seen you post or say anything about cardinal numbers. Do you have proof?"

The proof that the cardinality of the reals is greater than that of the algebraic numbers is a well-known proof given by Cantor. The proof itself is referred to as the diagonalization proof whereby (expressed in decimal form), you go down diagonally through the list of algebraic numbers to come up with a new number which isn't part of the list (which you already know is referred to as C for the reals).

That is not what Cantor's diagonal slash proves. What it does prove is that the reals are not countable. You still need to show that the algebraic numbers are countable.

CB

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