# Rudin functional analysis, theorem 2.11 (Open mapping theorem)

Going through the proof of such theorem, there are few bits I don't understand. I'll write down the proof and in in the middle I'll add some comments.

Statement:

Suppose

(a) X is an F-space,

(b) Y is a topological vector space

(c) $$\Lambda : X \to Y$$ is continuous and linear, and

(d) $$\Lambda(X)$$ is of the second category in $$Y$$

Then

(i) $$\Lambda(X) = Y$$

(ii) $$\Lambda$$ is an open mapping, and

(iii) $$Y$$ is an F-space

Proof

Note that (ii) implies (i), since $$Y$$ is the only open subspace of $$Y$$. To prove (ii), let $$V$$ be a neighborhood of $$0$$ in $$X$$. We have to show that $$\Lambda(V)$$ contains a neighborhood of $$0$$ in $$Y$$.

Why is that? Shouldn't be goal to prove that $$\Lambda(V)$$ is a neighborhood of $$0$$ in $$Y$$? When the author says "contains" I could picture a situation where $$\Lambda(V)$$ might be closed set with no empty interior, which would contain an open set, but this is isn't the definition of open mapping, so I'm confused.

Let $$d$$ be an invariant metric on $$X$$ that is compatible with the topology of $$X$$. Define $$V_n = \left\{x : d(x,0) < 2^{-n}r \right\}, n = 0,1,...$$ where $$r>0$$ is so small that $$V_0 \in V$$. We will prove that some neighborhood $$W$$ of $$0$$ in $$Y$$ satisfies $$W \subset \overline{\Lambda(V_1)} \subset \Lambda(V)$$ Since $$V_2 - V_2 \subset V_1$$ statement (b) of theorem 1.13 implies $$\overline{\Lambda(V_2)} - \overline{\Lambda(V_2)} \subset \overline{\Lambda(V_1)}$$

I guess $$V_2 - V_2 \subset V_1$$ because if $$x \in V_2 - V_2$$ then there are $$y,z \in V_2$$ such that $$x = y - z$$

And we have

$$d(x,0) = d(y - z,0) < d(y,0) + d(-z,0) = d(y,0) + d(z,0) < (2^{-2} + 2^{-2})r = 2^{-1}r$$

right?

The rest of the construction to prove (ii) seems clear to me. Later to prove (iii) the following $$f : X/N \to Y$$ is defined

$$f(x + N) = \Lambda x$$

why is such map an isomorphism? The remaining part to prove the homeomorphism seems clear.

For the first question first recall that a subset $$U$$ of a topological space is open if and only if every $$x\in U$$ has a neighborhood $$U_x$$ contained in $$U$$ (Proof: $$U=\bigcup_{x\in U}U_x^\circ$$ is the union of open sets).
So back to the context of the open mapping theorem. For $$x\in V$$ you have to show that $$\Lambda(V)$$ contains a neighborhood of $$\Lambda(x)$$. Let $$U$$ be a neighborhood of $$0$$ in $$X$$ such that $$x+U\subset V$$. According to Rudin's proof, $$\Lambda(U)$$ contains a neighborhood $$W$$ of zero in $$Y$$. Then $$\Lambda(x)+W$$ is a neighborhood of $$\Lambda(x)$$ and $$\Lambda(x)+W\subset \Lambda(x)+\Lambda(U)=\Lambda(x+U)\subset \Lambda(V).$$
The second part is ok (i don't see how you go from $$d(y-z,0)$$ to $$d(y,0)+d(-z,0)$$ without passing through $$d(y,0)+d(z,0)$$, but it is certainly correct).
The last part is standard. The map $$f$$ is surjective because $$\Lambda$$ is (see (i)) and injective because you factor out the kernel (if $$\Lambda x=0$$, then $$x\in N$$ and hence $$x+N=0+N$$).