Help with Rudin rank theorem proof! I am struggling through Rudin's proof of the rank theorem (9.32) in the baby Rudin book. There is a part in the proof where he claims that for a finite-dimensional linear operator A, if the set V is open, then A(V) is an open subset of the range of A. I have seem things about the open mapping theorem involving Banach spaces, but I am not on that level yet and I don't see why the justification of this statement could possibly involve Banach spaces, considering this book does not talk about those. How does Rudin justify this statement, at the level of this book? 

Thanks!
Pick any $x_0 \in V$. We will show that $Ax_0 $ is an interior point of $A(V)$.

By translating (i.e. consider $V - x_0$ instead of $V$), we can assume $x_0 = 0$.

Let $y_1, \dots, y_n$ be a basis of $\rm{Range}(A)$ and choose $x_1, \dots, x_n$ with $y_i = Ax_i$ for each $i$.

As $V$ is open with $0 \in V$, there is some $\varepsilon &gt; 0$ such that $\sum_i \alpha_i x_i \in V$ holds for all $\alpha_1, \dots, \alpha_n$ with $|\alpha_i| &lt; \varepsilon$ for all $i$.

This implies that $A(V)$ contains the set

$$
\bigg\{ \sum_i \alpha_i y_i \mid |\alpha_1|, \dots, |\alpha_n| &lt; \varepsilon\bigg\}.
$$

Why does that imply your claim?
This appears to be true only when range of A have dimension larger than 0. Otherwise, $A(V)=0$, which is not open.