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.