Finite Ordinal Plus Transfinite Ordinal

Theorem
Let $n$ be a member of the minimal infinite successor set (a set like the natural numbers).

Let $x$ be an ordinal that contains the minimal infinite successor set.

Then:
 * $n + x = x$

That is:
 * $\left({n \in \omega \land \omega \subseteq x}\right) \implies n + x = x$

where $\omega$ is the minimal infinite successor set.

Proof
By Transfinite Induction on $x$.

The proof will use $<$, $\in$, and $\subset$ interchangeably. This is justified by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.

Base Case
By our hypothesis, $\omega \le x$, so $x \not < \omega$, so we may begin our induction at $\omega$.

From these conclusions, we may deduce that:
 * $\displaystyle \omega = \bigcup_{y \mathop \in \omega} \left({n + y}\right)$