(@Arthur: Thanks very much. For the part $4 \Rightarrow 1$, I've tried to prove. But I'm not sure I'm right. So I write here as an answer, for the comment cannot contain the long proof.)
The part $(4 \Rightarrow 1)$
For $\delta=1$, it is obviously right, because the sum of any two finite ordinals is less then $\omega$.
Now assuming that for any $x<\delta$ the case is always right.
Suppose that $\delta$ is a successor. Let $\delta=y+1$. Then $\omega^\delta=\omega^y \times \omega$. $\beta$ and $\gamma$ must be bigger than $\omega^y$, otherwise $\beta+\gamma\le\omega^y$. So let $\beta=\omega^y \times i +j$, and $\gamma=\omega^y \times m + n$, where $i,j,m,n \in \omega$. Therefore, $\beta + \gamma\le \omega^y \times (i+1+m+1)\le \omega^\delta.$
Suppose that $\delta$ is a limit ordinal. Let $\omega^\delta=\sup\{\omega^\xi: \xi \in \delta\}$. Therefore for any $\beta, \gamma < \omega^\delta$, there exists a $\xi'$ such that $\beta, \gamma<\omega^{\xi'}$, and hence by the assuming, we have $\beta+\gamma<\omega^{\xi'}$.