# order type of club - cofinality

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.

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.)