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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.".

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    $\begingroup$ What is your definition of norm equivalence? $\endgroup$ Jun 10, 2014 at 7:16
  • $\begingroup$ If there exist two constant $K>0$ and $M>0$ such that $K||x||_a\le ||x||_b\le M||x||_a$ holds for each $x\in X$. @ChristopherA.Wong $\endgroup$
    – meysam
    Jun 10, 2014 at 7:18
  • $\begingroup$ And, what is your definition of generating the same open sets (same topology)? Specifically, what does it mean for a norm to generate open sets? $\endgroup$ Jun 10, 2014 at 7:23
  • $\begingroup$ With each norm we can define a meter: $d(x,y) = ||x-y||$, thus we have a metric space $(X,d)$ and in each metric space we have Opened and Closed set. @SammyBlack $\endgroup$
    – meysam
    Jun 10, 2014 at 8:12

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

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