Ordinal Addition is Left Cancellable
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Theorem
Let $x, y, z$ be ordinals.
Then:
- $\paren {z + x} = \paren {z + y} \implies x = y$
That is, ordinal addition is left cancellable.
Proof
For the proof, $<$, $\in$, and $\subsetneq$ will be used interchangeably.
This is justified by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal.
Note that:
\(\ds x < y\) | \(\implies\) | \(\ds \paren {z + x} < \paren {z + y}\) | Membership is Left Compatible with Ordinal Addition | |||||||||||
\(\ds y < x\) | \(\implies\) | \(\ds \paren {z + y} < \paren {z + x}\) | Membership is Left Compatible with Ordinal Addition |
However:
\(\ds \paren {z + x} = \paren {z + y}\) | \(\implies\) | \(\ds \paren {z + x} \not < \paren {z + y} \land \paren {z + y} \not < \paren {z + x}\) | No Membership Loops |
This contradicts the consequents of the first two equations, so:
\(\ds \) | \(\implies\) | \(\ds x \not < y \land y \not < x\) | Rule of Transposition | |||||||||||
\(\ds \) | \(\implies\) | \(\ds x = y\) | Ordinal Membership is Trichotomy |
$\blacksquare$
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Ordinal addition is not right cancellable.
By Finite Ordinal Plus Transfinite Ordinal
- $\paren {1 + \omega} = \paren {0 + \omega}$
but $1 \ne 0$.
Also see
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 8.5$