Open Mapping Lemma

Let and be Normed Space with Complete. Let be a Linear Operator. Suppose . Then for all and any fixed there is some with and for any fixed . In particular, is surjective.

Furthermore, is Complete.

Proof

Given , , seek with and (we took but proof is the same for any ) We know dense in , so there is some in with . Also is dense in , so there is some in with with . Continue obtaining . Put which converges as its a Cauchy Sequence.

Then and so

Now we prove is Complete. Given Cauchy in WLOG for all For each , choose with with Also choose with

Thus

Let (Converges as its a Cauchy Sequence)

Then

so is convergent, so is Complete.