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I'm using the definitions from Rudins book Functional Analysis.

Suppose $X,Y$ are topological vector spaces, $\Lambda: X \to Y$ is continuous and linear and $\Lambda(X)$ is of the second category in Y.

The proposition is that $\Lambda$ is open (the actual proof requires X to be an F-space).

I came up with the following proof, which should be incorrect since I don't use the invariant complete metric on X, but I can't see where I go wrong.

Let $V$ be a neighborhood of $0$ in $X$. Choose a balanced neighbourhood of zero $U$ in $X$ such that $\bar{U} + \bar{U} \subset V$. (This is always possible, and used by Rudin, eg. in his proof of the Banach-Steinhaus theorem).

Now, $X = \cup _{n\in\mathbb{N}} n\bar{U}$, and since $\Lambda(X)$ is of the second category and $\Lambda(\cup _{n\in\mathbb{N}}n\bar{U}) \subset \cup_{n\in\mathbb{N}} n\Lambda(\bar{U})$ the latter union is of the second category, which implies that $\Lambda(\bar{U})$ is of the second category.

This implies $\exists y\in \Lambda(\bar{U})$ and an open neighbourhood $W$ of zero in $Y$ such that $y+W \subset \Lambda(\bar{U}) \implies W \subset\Lambda(\bar{U}) - y \subset \Lambda(\bar{U}) - \Lambda(\bar{U}) = \Lambda(\bar{U} - \bar{U}) \subset \Lambda(V)$.

Since $V$ was arbitrary, this proofs that $\Lambda$ is open at the origin, and therefore $\Lambda$ is open.

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  • $\begingroup$ I don't think it's true that a subset of a topological vector space of the second category always has nonempty interior. $\endgroup$ – Rupert Dec 8 '16 at 13:22
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    $\begingroup$ $\overline{\Lambda(\overline{U})}$ has nonempty interior, but $\Lambda(\overline{U})$ need not be closed. $\endgroup$ – Daniel Fischer Dec 8 '16 at 13:23
  • $\begingroup$ yes that's it, stupid of me, thank you! $\endgroup$ – Scipio Dec 8 '16 at 13:25
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As pointed out by Daniel Fischer, the existence of $W$ is only guaranteed if $\Lambda(\bar{U})$ is closed, which in turn depends on $\Lambda$ being open..

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