Every well-ordered set is isomorphic to an ordinal. Common notion, for example see Introduction to https://arxiv.org/abs/2409.07352

⊢((𝐴∈V∧𝑅We𝐴)↔∃𝑓(dom𝑓∈On∧𝑓IsomE,𝑅(dom𝑓,𝐴)))

\[ \vdash ( ( A \in \mathrm{V} \wedge R \mathrm{We} A ) \leftrightarrow \exists f ( \mathrm{dom} f \in \mathrm{On} \wedge f \mathrm{Isom} \mathrm{E} , R ( \mathrm{dom} f , A ) ) ) \]

Every well-ordered set is isomorphic to
a unique ordinal.

⊢((𝐴∈V∧𝑅We𝐴)↔∃!𝑜∈On∃𝑓∈(𝐴↑ₘ𝑜)𝑓IsomE,𝑅(𝑜,𝐴))

\[ \vdash ( ( A \in \mathrm{V} \wedge R \mathrm{We} A ) \leftrightarrow \exists{!} o \in \mathrm{On} \exists f \in ( A \uparrow_\mathrm{m} o ) f \mathrm{Isom} \mathrm{E} , R ( o , A ) ) \]

We can phrase the Axiom of Choice as "Every set injects into an ordinal."

⊢(CHOICE↔∀𝑥∃𝑜∈On𝑥≼𝑜)

\[ \vdash ( \mathrm{CHOICE} \leftrightarrow \forall x \exists o \in \mathrm{On} x \preccurlyeq o ) \]

#math #metamath #SetTheory #WellOrdering #OrdinalNumbers

Here it is the 23th already, but the township zoning commission hasn't removed their sign.

#proofreading #OrdinalNumbers #EnglishIsHard

Nathan: 8, 9, 10, 11. There are 11 stars.
Me: Yeah. If we added another one, there would be 12. And if we took one away, there would be 10.
Nathan: If we took 10 away, there would be 9. If we took 9 away, there would be 8.

He continued this pattern until he got all the way to zero.

At first I was going to tell him that 10 minus 10 does not equal 9, but then I realized that he was pointing to individual stars and saying "10" to mean "the tenth one."

#tmwyk #OrdinalNumbers