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Suppose $f$ is a linear map between vector spaces, and whenever $U$ is an open set containing $0$, then $f(0)$ is an interior point of $f(U)$. Can we deduce that any open set containing $0$ has an open image by $f$? How?

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What does it mean for an arbitrary subset of a vector space to be open? – Michael Greinecker Sep 14 '12 at 10:46
I had "normed vector spaces" in mind, for the topology. – Tom Sep 14 '12 at 10:57
up vote 3 down vote accepted

Let $X$ and $Y$ be topological vector spaces and let $f\colon X\to Y$ be a linear function that takes zero neighbourhoods of $X$ into zero neighborhoods of $Y$.

Lemma: $f$ maps open sets in $X$ into open sets in $Y$.

Proof: Suppose that $N\subseteq X$ is an open set. Pick any $x\in N$. We will show that $f(x)$ is an interior point of $f(N)$. Notice that $N-x$ is a zero neighborhood. Thus, $f(N-x)$ is a zero neighborhood. This implies that $f(N-x)+f(x)$ is a neighbourhood of $f(x)$. Because $f$ is linear $f(N-x)+f(x)= f(N)$. We conclude that $f(x)$ is an interior point of $f(N)$. Because $x\in N$ was an arbitrary choice we conclude that $f(N)$ is open. QED

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