Q:

Open set generated by two equivalent norm

Suppose X is an vector space and $||.||_a$ and $||.||_b$ are two norm on it, how i can proof that ''This two norms are equivalent iff they generate same open sets."?
P.S.: Sense of question made by ''Principles of Real Analysis, C.D. Aliprantis, O. Burkinshaw, 3rd Edition, p220.".

A:

It is easy to see that equivalent norms yield the same open sets, because if $K \cdot \Vert x \Vert_a \leq \Vert x \Vert_b \leq M \Vert x \Vert_a$, you have
$$
B_r^{\Vert \cdot \Vert_a}(x_0) \subset B_{r \cdot M}^{\Vert \cdot \Vert_b}(x_0)
$$
and vice versa (show this!).
As a set $U$ is open (w.r.t. $\Vert \cdot \Vert$) iff for each $x \in U$ there is some $\varepsilon >0$ with $B_\varepsilon^{\Vert \cdot \Vert} (x) \subset U$, you should be able to show that $U$ is open w.r.t. $\Vert \cdot \Vert_a$ iff it is open w.r.t. the other norm.
For the converse, note that $B_1^{\Vert \cdot \Vert_a}(0)$ is open w.r.t. the first norm (why?), so by assumption also w.r.t. the second norm.
Thus, there is $\varepsilon > 0$ with
$$
B_\varepsilon^{\Vert \cdot \Vert_b}(0) \subset B_1^{\Vert \cdot \Vert_a}(0).
$$
This should allow you to conclude that $\Vert \cdot \Vert_a \leq \frac{1}{\varepsilon} \Vert \cdot \Vert_b$.

