# How to show $\operatorname{card}(\omega+1)=\omega$

Apparently $$\operatorname{card}(\omega+1)=\omega$$. This means that there is an order $$<$$ on $$\omega+1$$ such that there is an isomorphism of ordered set $$f$$, $$(\omega+1,<) \cong (\omega,\in)$$, but to what does $$f$$ send $$\omega \in \omega+1$$?

Edit

Also to show $$card(\omega+1)=\omega$$ I think I need to show that any ordinal $$\alpha$$ that is in bijection with $$\omega+1$$ is such that $$\alpha \ge \omega$$ no? [I know also that if two ordinals are isomorphic ($$\langle \beta, \in \rangle \cong \langle \gamma, \in \rangle$$ i.e. isomorphism of ordered set) then $$\beta=\gamma$$, is this usefull here?.]

If it it only concerns cardinality then a bijection is enough.

If you insist on an order preserving bijection then:

Let $$<$$ on $$\omega+1$$ be defined by $$\omega<0$$ and $$n if $$n,m\in\omega$$ with $$n\in m$$.

Then $$f$$ can be prescribed by:

• $$\omega\mapsto0$$
• $$n\mapsto n+1$$ for $$n\in\omega$$
• Thanks, I forgot one thing though that I edited in the question. – roi_saumon Jan 22 at 15:10

This is know as the Hilbert hotel paradox. The Hilbert hotel have an infinite number rooms (all aligned in one dimension), so we assimilate it to $$\omega$$.

$$1$$ new customer arrives, but the hotel is full, every room is occupied by $$\omega$$ customers.

But the butler just invites the new customer to take the first room (labelled $$1$$) and asks the initial occupant to move to the next room ($$1\mapsto 2$$) and so on ($$n\mapsto n+1$$) for successive occupants.

Finally $$\omega+1$$ customers can fit in the Hilbert hotel.