Proving nicely that $3+\omega=\omega$ where $\omega$ is the smallest limit ordinal.

Question

I just learned about ordinal numbers and I am a bit confused to show things with them. Let $\omega$ be the smallest limit ordinal. I want to show that $3+\omega=\omega$. So the definition I have of $3+\omega$ (since $\omega$ is a limit ordinal) is : $3+\omega = sup\{n+3:n<\omega \}$. Then the definition of the sup I have is $supX=\{\gamma : \exists x \in X \text{such that } \gamma \in x\}$, so we have : $3+\omega = \{\gamma : \exists x \in \{n+3:n<\omega \}\ \text{such that } \gamma \in x\}$. But how to show this is equal to $\omega$?

Answer 1

For any $n < \omega$, $3+n < \omega$, so $3+\omega \leq \omega$. On the other hand, for any $n < \omega$, $3+\omega \geq 3+n > n$. Thus $3+\omega \geq \omega$.