Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

1 Answer 1

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.