Knowlpedia

Basic properties of interior

Monotonicity, idempotence, and compatibility with finite intersections

Proposition. Let (X,d)(X,d) be a metric space and let A,B⊂XA,B\subset X. Then:

  1. If A⊂BA\subset B, then A∘⊂B∘A^\circ\subset B^\circ.
  2. A∘=AA^\circ=A if and only if AA is .
  3. (A∘)∘=A∘(A^\circ)^\circ=A^\circ (idempotence).
  4. (A∩B)∘=A∘∩B∘(A\cap B)^\circ=A^\circ\cap B^\circ.
Remarks

Proof sketch.

  1. Any open set contained in AA is contained in BB, so its union is contained in B∘B^\circ.
  2. By definition, A∘⊂AA^\circ\subset A always; equality holds exactly when AA is already open.
  3. A∘A^\circ is open, so its interior is itself.
  4. Use that a set is contained in A∩BA\cap B iff it is contained in both AA and BB.