Hi,
I'm not sure if this is the right section, but I'm talking about numbers .
The questions is as written in the title: Is a number preceding infinity, finite?
define "precede" and "infinity".
the reaon i make this request is this:
supposing you defined infinity to be the number of (distinct) real numbers, and defined "precede" to mean, any number strictly less than the number of real numbers.
the number of rational numbers is strictly less than the number of real numbers, but this number is not finite.
it turns out there are different "kinds" of infinity, and they act a bit differently. the "smallest" infinity is used often enough to get its own name, $\displaystyle \omega$.
this is the number of natural numbers (which turns out, strangely enough, to be the same number as the number of integers, AND the number of rational numbers. infinte things behave very strangely).
it's safe to say that if n < $\displaystyle \omega$, then n is finite (sometimes this is even written as n < $\displaystyle \infty$).
for a not-very-technical view of this, check out this video series:
Horizon - To infinity and Beyond part 1 of 6 - YouTube
So, in order for infinity to be infinite it must have infinite predecessors? Hence, anything preceding infinity is infinite? I suppose infinity should be treated differently since there is no border between something being finite and something being infinite. It is not as though I can say finiteN+1=infinity because unlike infinity both finiteN and 1 are finite.
Have I finally made sense of the situation?
OP,
I believe that you are asking a set theoretical question. The smallest level of infinity is w=|N| where N is the set of natural numbers. w has no immediate predecessor because n<w for each natural number n. However, there are proper subsets of N that have size w. For example, the set E of even natural numbers has size w. That is , there is a one-to-one correspondence between N and E, for example {(n,2n)|n in N}.