Ordinal Addition by Zero

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Theorem

Let $x$ be an ordinal.

Let $\varnothing$ be the zero ordinal.


Then:

$x + \varnothing = x = \varnothing + x$

where $+$ denotes ordinal addition.


Proof

By definition of ordinal addition, it is immediate that:

$x + \varnothing = x$

$\Box$


We shall use Transfinite Induction on $x$ to prove $\left({\varnothing + x}\right) = x$


Base Case

The induction basis $x = \varnothing$ comes down to:

$\varnothing + \varnothing = \varnothing$

This follows by the above.

$\Box$


Inductive Case

For the induction step, suppose that $\varnothing + x = x$.

Then, also:

\(\displaystyle x^+\) \(=\) \(\displaystyle \left({\varnothing + x}\right)^+\) Substitutivity of Equality
\(\displaystyle \) \(=\) \(\displaystyle \varnothing + x^+\) Definition of ordinal addition

$\Box$


Limit Case

Finally, the limit case.

So let $x$ be a limit ordinal, and suppose that:

$\forall y \in x: \varnothing + y = y$


Now we have:

\(\displaystyle x\) \(=\) \(\displaystyle \bigcup_{y \mathop \in x} y\) Union of Limit Ordinal
\(\displaystyle \) \(=\) \(\displaystyle \bigcup_{y \mathop \in x} \left({\varnothing + y}\right)\) Indexed Union Equality
\(\displaystyle \) \(=\) \(\displaystyle \varnothing + x\) Definition of ordinal addition

$\Box$


Hence the result, by Transfinite Induction.

$\blacksquare$


Sources