Countability and Cardinality

By Brad Rodgers on Saturday, November 10, 2001 - 05:38 am:

Is it true, and is it provable that given the set of algebraic numbers, and a set of finite transcendental numbers, it is possible to form the set of real numbers using only a coutably infinite numbers of operations?

Similarly, can all real numbers be formed by the operation

\sum_{n = 0}^{\infty} g (n)

for

g (n)

always being rational at integer

n

? (

g (n)

is different for different number being formed)
The former will follow from the latter (if the latter is indeed true).

Brad

By Dan Goodman on Saturday, November 10, 2001 - 03:46 pm:

Yup, you can form all real numbers from finite sums of rationals. In fact, this is one definition of the real numbers (well, sequences are usually used rather than sums). Suppose

x

is a real number, let

r_{n} =

the first

n

places in the decimal expansion of

x

, clearly a rational number. Now let

s_{0} = r_{0}

s_{n + 1} = r_{n + 1} - r_{n}

. So

\sum_{n = 0}^{N} s_{n} = r_{N}

. Clearly

r_{N} \to x

N \to \infty

(since

r_{N} < x

for all

N

and for any number

y < x

you can find an

N

with

r_{N} > y

, so

r_{N}

must tend to

x

) so

\sum_{n = 0}^{\infty} s_{n} = x

By Brad Rodgers on Saturday, November 10, 2001 - 07:45 pm:

Thanks. Here's another similar one that seems bound to be false, but is it possible to form all real numbers from a finite number of operations between all algrebraic numbers and a finite number of transcendental numbers? It seems as though the two sets would have different cardinality, but I can't prove this with any rigor.

Brad

By Dan Goodman on Saturday, November 10, 2001 - 07:55 pm:

You're right, they would have different cardinality. In fact, even if you allow a countably infinite number of operations but only a finite number of operations used in forming any particular number, then this will not include all the reals. The reason is that a countable union of countable sets is countable, and the reals are uncountable. Let A₀ ={the algebraic numbers and a countable number of transcendental numbers (note that this is slightly more than you started with)} and A_n+1 be the set formed by adding, multiplying or dividing by two numbers in A_n . Each A_n is clearly countable and the union of all the A_n gives you all the numbers which can be formed using a finite number of operations on algebraic and a countable number of transcendental numbers. However, it is a countable union (because it is indexed by natural numbers n) of countable sets. So it is a countable set.

By Dan Goodman on Saturday, November 10, 2001 - 07:58 pm:

By the way, now that I look at it, the first thing you posted was slightly misleading. In fact, doing a countable number of operations on what I called A₀ above (including infinite summations) would not form all reals, for the same reason as above (a countable number of operations on a countable set only gives a countable set). Even though you can form each real with a countable number of operations it is impossible to form all reals simultaneously with a countable number of operations.