# Implications of continuity on normed spaces

My goal for this question is to understand the implications of continuity its relationship to normed spaces. This question is based off of the notes for 18.155 available at ocw.mit.edu.

Let $u$ be a linear functional $u:V\rightarrow R$ on a normed space $V$. Let's assume we know that $u$ is continuous at $0$. This implies that $u^{-1}(-1, 1)$ is a neighborhood of $0 \in V$.

The text continues to say that this implies that $\exists \epsilon>0$ such that $u(\{f \in V: ||f|| < \epsilon\}) \subset (-1,1)$. Simply put, I don't understand why this is implied. So two questions: Why is this implied? Why does the value of the norm have anything to do with a neighborhood based on the distance function?

Any help at all would be appreciated. Thanks much!

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$\newcommand{\eps}{\epsilon}$A set $U$ in a metric space $(X,d)$ is open if for every $x \in U$ there exists an $\epsilon>0$ such that $B(x,\epsilon) \subset U$. A normed vector space is a metric space with metric given by $d(v,w)=||v-w||$. In your question we have that $U=u^{-1}((-1,1))$ is an open set containing $0$ thereby there is an $\epsilon>0$ such that $B(0,\eps) \subset U$ but
$$B(0,\eps)=\{f \in V : ||f-0|| < \eps\},$$
in particular $||f-0||=||f||$. Finally we see that since $B(0,\eps) \subset U$ so when we push forward under $u$ we must have that $u(B(0,\eps)) \subset u(U)=(-1,1)$.