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.



(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 $Y$


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

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

(iii) $Y$ is an F-space


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

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


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 $x+N=0+N$).

  • $\begingroup$ Hi, I know this post is a bit old. But I was wondering one thing since I'm going through this stuff again. In the proof it is explictly mentioned that $X/N$ is an F-space, and therefore we have to prove that there's an isomorphism which is also a homeomorphism. Why exactly do we need such features? Also why not trying to do the same with $X$ since it is an F-space already. $\endgroup$ May 20 '19 at 16:32

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