If A contains all its accumulation points, then A is closed
We have a set A that contains all its accumulation points. We want to prove A is closed, which means proving Aᶜ (the complement) is open.
1. a ∈ Aᶜ → a ∉ A
2. A contains all accumulation points + a ∉ A → a is not an accumulation point
3. a not accumulation point → ∃ε: neighborhood misses A (except possibly at a)
4. a ∉ A + neighborhood misses A → entire neighborhood ⊆ Aᶜ
5. Every point of Aᶜ has such a neighborhood → Aᶜ is open → A is closed ∎
(⇒) A closed → contains acc. pts.
Proof by contradiction: if acc. pt. a ∉ A, then a ∈ Aᶜ (open), so some neighborhood misses A entirely — contradiction.
(⇐) Contains acc. pts. → A closed
Direct proof: show Aᶜ is open. Any a ∈ Aᶜ isn't an acc. pt., so some neighborhood misses A, hence lies in Aᶜ.