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.".

  • 2
    $\begingroup$ 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

1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.