How to write $pi$ as a set in ZF?

I know that from ZF we can construct some sets in a beautiful form obtaining the desired properties that we expect to have these sets. In ZF all is a set (including numbers, elements, functions, relations, etc...).

For example we can define a copy of $mathbb{N}$, using the empty set axiom and others, called sometimes $omega$ and defining,

begin{align}
0_mathbb{N}&={}\
1_mathbb{N}&={{}}\
2_mathbb{N}&={{},{{}}}\
3_mathbb{N}&={{},{{}},{{},{{}}}}\
4_mathbb{N}&={{},{{}},{{},{{}}}, {{},{{}},{{},{{}}}}}\
& vdots
end{align}

Well, ZF allow us to build this type of sets that are kind of arrangements of brackets and commas. We can follow with $mathbb{Z}$ since we can define $(a,b)={{a},{a,b}}$:

$mathbb{Z}$ is defined as the set of equivalence classes $mathbb{Z}=(mathbb{N}timesmathbb{N})big/sim$ where

$$sim, ={big((m,n),(h,k)big)in(mathbb{N}timesmathbb{N})times (mathbb{N}timesmathbb{N}):(m+_mathbb{N} k)= (h+_mathbb{N} n)}$$

here, the integers are sets more complicated than natural numbers. For example,

begin{align}
-2_mathbb{Z}&={(1_mathbb{N},3_mathbb{N}),(2_mathbb{N},4_mathbb{N}),(3_mathbb{N},5_mathbb{N}),ldots,(n_mathbb{N},(n+2)_mathbb{N}),ldots}\
&={{{1_mathbb{N}},{1_mathbb{N},3_mathbb{N}}},{{2_mathbb{N}},{2_mathbb{N},4_mathbb{N}}},ldots}\
-2_mathbb{Z}&={{{{{}}},{{{}},{{},{{}},{{},{{}}}}}},\
&quad{{{{},{{}}}},{{{},{{}}},{{},{{}},{{},{{}}}, {{},{{}},{{},{{}}}}}}},ldots}\
end{align}

Here we note the importance of notations. We continue with $mathbb{Q}=(mathbb{Z}times(mathbb{Zsetminus{0_mathbb{Z}})}})big/sim$ where

$$sim, ={big((m,n),(h,k)big)in(mathbb{Z}times(mathbb{Zsetminus{0_mathbb{Z}})}})times (mathbb{Z}times(mathbb{Zsetminus{0_mathbb{Z}})}}):modot_mathbb{Z}k=hodot_mathbb{Z}n}$$

And for example:

$$(0.2)_mathbb{Q}={(1_mathbb{Z},5_mathbb{Z}),(2_mathbb{Z},10_mathbb{Z}),(3_mathbb{Z},15_mathbb{Z}),ldots,(n_mathbb{Z},(5n)_mathbb{Z}),ldots}$$

Imagine if we write the integer numbers as before and write the ordered pairs in unabridged form ($(0.2)_mathbb{Q}$ is a nice abbreviation for this monster). However we can.

Finally we define $mathbb{R}$ as the set of all Dedekind cuts, for example:

$$(0.2)_mathbb{R}={xinmathbb{Q}:x<_mathbb{Q} (0.2)_mathbb{Q}}$$

Note that $(0.2)_mathbb{R}$ is even more monstrous than $(0.2)_mathbb{Q}$. Also I can write $(sqrt{2})_mathbb{R}$ showing its elements in a simple form,

$$(sqrt{2})_mathbb{R}={xinmathbb{Q}:(x^2<_mathbb{Q} 2_mathbb{Q}) lor (x<_mathbb{Q} 0_mathbb{Q})}$$

But I don't know how to do it with $pi_mathbb{R}$, since

$$pi=lim_{ktoinfty}sum_{n=0}^{k}cfrac{2^{n+1} n!^2}{(2n + 1)!}$$

I only know that

$$(pi)_mathbb{R}=bigcup_{k=1}^{infty} left(sum_{n=0}^{k}cfrac{2^{n+1} n!^2}{(2n + 1)!}right)_mathbb{R}$$

Since monotonically converge to $pi$ we have

$$(pi)_mathbb{R}=bigcup_{k=1}^{infty} left{xinmathbb{Q}:x<_mathbb{Q} left(sum_{n=0}^{k}cfrac{2^{n+1} n!^2}{(2n + 1)!}right)_mathbb{Q}right}$$

Is there any way to avoid infinite union (and
the choice of a particular convergent sequence
) as the case of $(sqrt{2})_mathbb{R}$? If not, why?

Can we write the set representing $pi$ listing its elements as we do with integers or rationals (as $mathbb{Q}$ is countable I guess that should be able to do, but I don't know how to do)?

If in ZF all is a set, is so surprising the fact that so many things can be defined, then my last question is

How many more things can be built using ZF?, ZF could define us what is a derivative, an integral, a limit or a measure?

Thanks in advance.

edited May 24 '13 at 7:36

asked May 24 '13 at 7:21

Gaston Burrull

2,80732056

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Possible duplicate: math.stackexchange.com/q/62852/8087
$endgroup$
– Alexander Thumm
May 24 '13 at 8:20

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@AlexanderThumm I'm not asking about integers. Answers in that topic are quite different from those that got here and similar to what I explained before my questions
$endgroup$
– Gaston Burrull
May 25 '13 at 3:27

1

$begingroup$
My bad. At least the answers are mildly related so the other question should still be of interest.
$endgroup$
– Alexander Thumm
May 25 '13 at 16:58

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