← Back to Home

Lemma 2.3.2 — Visual Proof

If A is closed, then A contains all its accumulation points

A (closed set)a(accumulation point)
1

Setup

A is a closed set (shown in blue). We have a point 'a' that is an accumulation point of A.

💡 Key Insight

The contradiction arises from two incompatible requirements: being an accumulation point means every neighborhood must intersect A, but being in an open set (Aᶜ) means some neighborhood is entirely contained in Aᶜ and thus misses A completely.