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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.".
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 $\Vert \cdot \Vert_a \leq \frac{1}{\varepsilon} \Vert \cdot \Vert_b$.
