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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?.]

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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 $n<m$ if $n,m\in\omega$ with $n\in m$.

Then $f$ can be prescribed by:

  • $\omega\mapsto0$
  • $n\mapsto n+1$ for $n\in\omega$
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  • $\begingroup$ Thanks, I forgot one thing though that I edited in the question. $\endgroup$ – roi_saumon Jan 22 at 15:10
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This is know as the Hilbert hotel paradox.

enter image description here

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.

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