Ordinal Addition/Examples/Ordinal Addition by Two

Theorem

Let $x$ be an ordinal.

Let $x^+$ denote the successor of $x$.

Let $2$ denote the successor of the ordinal $1$.

Then:

$x + 2 = x^{++}$

where $+$ denotes ordinal addition.

Proof

\(\ds x + 2\)

\(=\)

\(\ds x + 1^+\)

\(\ds \)

\(=\)

\(\ds \paren {x + 1}^+\)

Definition of Ordinal Addition

\(\ds \)

\(=\)

\(\ds x^{++}\)

Ordinal Addition by One

$\blacksquare$

Sources

• 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $5$: Ordinal Numbers: $\S 3$ Some ordinals