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 $\aleph_a$.