I'm starting reading about cardinals (Shoenfield, page 253) and showing $x \subseteq y \implies \alpha = Card(x) \le Card(y) = \beta$. I notice that there is a injection $\alpha \to \beta$, so I want to conclude $\alpha \subseteq \beta$. Injection between ordinals $\alpha \hookrightarrow \beta$ asks the same but with an order-preserving injection. So maybe this property is false. I was wondering what is an easy counterexample.
My try
I examine first what can be the proof of Injection between ordinals $\alpha \hookrightarrow \beta$. As shown there, we can built an order preserving injection $h:\beta \to \beta$. Therefore, $\beta \cong h(\beta)$. Now, I'm proving that $h(\beta)$, which is clearly a set of ordinals (using axiom of replacement), is transitive. I believe $x \in h(\beta) \implies h^{-1}(x) \in \beta \implies h^{-1}(x) \subseteq \beta \implies x \subseteq h(\beta)$ where I use transitivity of $\beta$. So, since every transitive set of ordinals is an ordinal, we conclude $h(\beta)$ is an ordinal. But $\beta \cong h(\beta)$ implies $\beta = h(\beta)$ which contradicts the assumption $h(\beta) \subset \beta$ (since $\beta \subset \alpha$ so there are elements of $\alpha$ that are not mapped in the image.
So we see that it was needed to have an isomorphism, not just an injection.
Best Answer
There is an injection $f: \omega + 1 \to \omega$ defined by $f(\omega)=0, f(n)=n+1$ for $n \in \omega$, but it's not the case that $\omega \cup \{ \omega \} = \omega + 1 \subseteq \omega$. More generally, there's an injection from any countable ordinal into $\omega$. That's how you know the ordinal is countable.