I have an ordinal $\alpha$ and an infinite cardinal $\mathfrak{a}$ and I want to show that $\alpha \in \mathfrak{a}$ implies $\alpha + \mathfrak{a} =\mathfrak{a}$ using ordinal addition, as in $0+\omega=\omega=3+\omega\neq \omega+3$, to make the notation clear.
I'm still very bad with limit ordinals in the definition of ordinal addition, so I don't know how to approach this.
(1) Let ${\frak a}$ be cardinal, then let $x,y∈{\frak a}$ be 2 ordinals, by definition of cardinals, $|x|,|y|<{\frak a}$, and so, because $|x+y|<{\frak a}$, we also have $x+y∈{\frak a}$.
${\frak a}$ is a limit ordinal, so $a+{\frak a}=\sup\{a+b\mid b\in{\frak a}\}$, by (1) we have that for all $z ∈\{a+b\mid b∈{\frak a}\}, z\in{\frak a}$, so $a+{\frak a}={\frak a}$ or $a+{\frak a}∈{\frak a}$, if $a+{\frak a}∈{\frak a}$, then $a+{\frak a}+1∈{\frak a}$, and so $a+{\frak a}∈a+a+{\frak a}+1\in\{a+b\mid b ∈{\frak a}\}\implies a+{\frak a}∈\sup\{a+b\mid b\in{\frak a}\}$. Which is absurd, so $a+{\frak a}={\frak a}$