Let me try this way: a number in R is a sum of infinite serie:  sum (n 1 ->
oo) (a_n)/10^n, where all a_n, which is a placeholder, have some integer in
the set {0,1,2,3,4,5,6,7,8,9}. Therefore we may express all the numbers in R
[0,1), if we have all possible combinations of a_n.
The infinite integer N(inf) is analogically: sum (n 0 -> oo) (a_n)10^n,
where all a_n, which is a placeholder, have some integer in the set
{0,1,2,3,4,5,6,7,8,9}.

The integer is finite, and therefore a special case of N(inf), if all
placeholders beginning from some arbitrary placehoder n>1 have filled by
zeros, i.e. on the left side placeholders.
Therefore we may express all the infinite integers in N(inf) [0,omega), if
we have all possible combinations of a_n.

Now, for every N(inf) [0,omega) there is some real in R [0,1) and vice
versa. This is one-to-one mapping. (Note finite integer 1 maps to omega -
above). Therefore it is also a clear consequence that there is no one-to-one
mapping from finite N [0,omega) to R [0,1).

The bijection is more evident if you express integers in N(inf) as "decimal
numbers" when the point of reference is omega zero instead of  standard
zero. An example that hopefully clarifies this is the following:
Assume we have an infinite integer ...999 (or ...9). This is the infinite
sum (n 0->oo) (a_n)10^n, where all (or every) a_n equals to 9. It is
important to notice that the sum must result in ...9. If the sum does not
result in that infinite integer, then the sum must wrong. Usually the
infinite sum is defined as a limit, but we do not search the limit of that
sum! We search the sum instead of it's limit. The sum is correct, if it (the
sum) really results in the sum instead of limit.
Because all integers have a successor, we must be always able to add one.
But now all the possible placeholders are already occupied with the greatest
possible member of the set {0,1,2,3,4,5,6,7,8,9}. The next number must be
greater than any infinite integer N(inf). Therefore ...9 +1 or written in
the infinite form(at) ...9 + ...01 = omega. Thus, directly according to the
definition of omega. Therefore also ...9 is the greatest possible infinite
integer < omega. Note that ...9 is odd. If we add 1 (or ...01) omega must be
even. But omega is even only from the point of view that is the standard
point of reference, i.e. standard zero = the normal zero.  Omega is the
first number one "behind" or "above" of N(inf), i.e. the first transinfinite
number. (Normally called transfinite). Because omega is the first
"transfinite number one", it  is odd as we consider even and odd numbers
from the omega point of view.
Now let´s describe omega zero (0). Omega zero is the point of view that is
infinite far from standard zero point of view. The word "far" is used here
only as an illustrative meaning. If we write the infinite integer ...9 as a
"decimal number" when the point of reference is omega, then
...9 = 0,9...  (or 0,999...) Note 0, is here omega zero reference point.
Note: we had to add ...01 or simply finite 1 to get omega.
Thus ...9 <omega. And ...9 divided by omega results in 0,999... Therefore
the infinite sum  (n 0->oo) (a_n)10^n, where all (or every) a_n equals to 9,
results in ...9 but the limit of the the infinite sum (n 0->oo) (a_n)10^n,
where all (or every) a_n equals to 9, results in omega. Similarly 0.999...
<1 but the limit of 0.999... written as a infinite sum equals to 1.
If we devide ...9 by omega we will receive 0,999...

It follows infinite integers N(inf) [0,omega) and reals R [0,1) are well
ordered. The smallest member in N(inf) is finite 1 or (...01 written in the
infinite format) and the smallest member in R [0,1) is what we call epsilon,
which has digit 1 in a placeholder a_n as n ->oo.
It also follows: oo or infinity is not a number, which is consistent with
the standard theory. Numbers are digit strings in one dimension. oo
describes the "distance" of placeholders a_n in that string, if I somehow
want to describe the property of infinity in the dense placeholder line.
Actually - you hopefully observe - infinity is the concept relatated to the
point of reference. (Therefore I may say point of references are nested like
the peals of onion.)
By changing the point of reference you can always write any digit string R,
N(inf), Omega N(inf) etc. like reals or infinite integers. We call those
numbers by different name because we have a different point of reference,
i.e. "the decimal point". As an example omega zero point something
(0,xxx...) or standard zero dot something (0.xxx...). It follows- if we do
not specify the point of reference, then the presentations or expressions
are "formally" identical (compare - bijection in the case of 0.999... and
0,999... =...9).

If you now consider reals R [0,1) and N(inf) [0,omega) it should be evident
that there exist one-to-one mapping between N(inf) [0,omega) and R [0,1),
i.e. bijection.

The order - iff you mean the successor - is of course standard and normal.
Please - remember  finite integers have non-significant zeros on the left
hand side of digit string. The order (what is the successor, which is
greater or smaller) will be determined by the difference. ...2 >...1,
because the difference equals to ...1.  ...02 >...01, because the difference
equals to 1 etc.  The successor of an integer y is x , iff x-y=1, i.e.
y+1=x.

Are there more odd than even numbers (including also infinite integers <
omega) ?
(Note: infinite integers are here rigorously defined as a sum of infinite
serie, analogous to reals (R), use dejagoogle for details)

Tapio"

The end of copy

I wish I been for your help!

A Hint: We have noticed above that the first odd integer is 1 or ...01
written in the infinite format. The greatest possible infinite integer
<omega equals to ...9 that is odd.