order type of club - cofinality

Question

With $ot(C)$ being the order-type of the club $C$, and $cof(\alpha)$ the cofinality of the ordinal $\alpha$

Show that for every limit ordinal $\alpha$ there is a club $C\subseteq \alpha$ such that $ot(C)=cof(\alpha)$

The whole notion of cofinality is new to me.

I guess that I need to define $X_i$ a cofinal sequence of $\alpha$ in $C$, then for all $i$, $ot(X_i)=cof(\alpha)$ by bijection between $\alpha$ and $cof(\alpha)$. And then i would need to take a closure of the sequence, and proving that the order type of the closure is still $cof(\alpha)$.

I'm really fuzzy about the details though.

Answer 1

Hint: Construct a strictly increasing, cofinal, continuous function $f \colon \mathrm{cof}(\alpha) \to \alpha$.

Then $\mathrm{ran}(f) \subseteq \alpha$ is a club of order-type $\alpha$.

Also note that order-types are not preserved by bijections -- they're only preserved by order-isomorphisms. (For instance: There is a bijection between $\omega$ and $\omega + \omega$ but they represent different order-types.)