Question

Let α be an ordinal. Show that |α| ∈ α.

Answer 1

Answer:

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Down vote

Let α be an ordinal and A be a set of ordinals. Then supβ∈A(α+β)=α+supβ∈A(β)

proof-explanation ordinals

My idea is to prove that α+supβ∈A(β) is the supremum of {α+β∣β∈A}. While I'm able to prove that α+supβ∈A(β) is a upper bound of {α+β∣β∈A}, I failed to show that α+supβ∈A(β)≤γ where γ is an upper bound of {α+β∣β∈A}.

The addition of ordinals is defined as follows α+0=α, α+(β+1)=(α+β)+1, and α+β=supγ<β(α+γ) if β is limit.

My attempt:

For all β∈A,β≤supβ∈A(β), then α+β≤α+supβ∈A(β) and thus α+supβ∈A(β) is a upper bound of {α+β∣β∈A}. Assume that γ is an upper bound of {α+β∣β∈A}, then α+β≤γ for all β∈A.