I Why is cf(x) not equal to \aleph_a when x=\aleph_{a} and a is a limit ordinal?

[cragar]

if x= \aleph_{a} where a is a limit ordinal. then cf(x)=cf(a)
why is the cf(x) not eqaul to \aleph_{a}
is it constructing an order type from the previous cardinals, and using the previous cardinals to construct a sequence

 

[stevendaryl]

You have to go back to the definition of cofinality of an ordinal. It's slightly confusing, but here's what I think the definition amounts to:
(I'm going to assume the axiom of choice, because everything gets messier without it)

Let A be an ordinal (which we can represent as the set of all smaller ordinals). Let B be a proper subset of A (that is, B is a set of ordinals, all of which are smaller than A). Then B is cofinal in A if for every \alpha < A, there is a \beta \epsilon B such that \alpha \leq \beta. In other words, B contains arbitrarily large elements of A. So the definition of the cofinality of A: It's the smallest cardinal \alpha such that there is a set B of size \alpha that is cofinal in A.

So a couple of examples: If n is finite ordinal greater than zero, then the cofinality of n is 1. That's because we can let B just be the one-element set B = \{ n-1 \}: If n' < n, then n' \leq n-1.

Another example is \omega: the cofinality of \omega is \omega. To see that, let B be any finite set of natural numbers. Then it has a largest element, max(B). Clearly, this number can't be greater than or equal to every element of \omega. So B is not cofinal in \omega. Turning that around, if B IS cofinal in \omega, then B must be infinite, so its cardinality is \omega.

So now, let's look at the case of \aleph_a. If a is a limit ordinal, then we can let B = \{ \aleph_{a'} | a' < a \}. Then B will be cofinal in \aleph_a. So the cofinality of \aleph_a would be less than or equal to the cardinality of B, which is just a.

 

[economicsnerd]

And notice that stevendaryl's argument that $cf(\aleph_a) \leq |a|$ can be slightly modified to show $cf(\aleph_a) \leq cf(a)$. Indeed, letting $C$ be some cofinal subset of $a$, one can verify that $\{\aleph_c: \ c\in C\}$ is cofinal in $\aleph_a$.