Assume that $V$ is a vector space and two norms $||.||$ and $||.||'$ are defined on $V$. We say $||.||$ and $||.||'$ are equivalent if there exist $M,m \gt 0$ such that:
$\forall x \in V \space\space m||x||\le ||x||' \le M||x||$


Assume that $V$ is an arbitrary vector space (not necessarily of finite dimension) and $||.||$ and $||.||'$ are two equivalent norms on it.

Prove that $A \subseteq V$ is open with respect to $||.||$ iff its open with respect to $||.||'$. Finally, Conclude that the topology induced by $||.||$ is the same as the one induced by $||.||'$.

My problem:

I don't know what a topology is and how to show that a set is open. Our teacher didn't define these two and assumed that we've learned them in other courses. So, I can't even get started. Any idea? help?

  • 1
    $\begingroup$ If you don't know what topology is, then you need to learn topology before you answer this question... $\endgroup$
    – 5xum
    Apr 13, 2017 at 9:02
  • $\begingroup$ Are you referring to the induced metric topologies? $\endgroup$ Apr 13, 2017 at 9:06
  • $\begingroup$ @ΘΣΦGenSan I think that's true... $\endgroup$ Apr 13, 2017 at 9:10
  • $\begingroup$ @5xum I Agree with you... but anyway, this is my situation. Isn't there a way to briefly learn just what is needed for solving this question? If u can, please just answer the question and i'll learn whatever is needed for understanding your answer. I don't want a topology course written in an answer :) $\endgroup$ Apr 13, 2017 at 9:12
  • $\begingroup$ @ArmanMalekzade You might not be familiar with topologies, but you are with "open sets", right? If not then a part of my answer can be looked at as a definition of open sets. From now on just keep in mind that the topology on a space is just the collection of open sets. In my answer I avoid the term topology delibaretely. $\endgroup$
    – drhab
    Apr 13, 2017 at 9:32

1 Answer 1


$A$ is open with respect to $||.||$ if for every $a\in A$ there is some $r_a>0$ such that $\{x\mid ||x-a||<r_a\}\subseteq A$.

Now observe that $||x-a||'<r_am\implies ||x-a||<r_a$ so that

$\{x\mid ||x-a||'<mr_a\}\subseteq\{x\mid ||x-a||<r_a\}\subseteq A$ for every $a\in A$.

And we conclude that $A$ is open with respect to $||.||'$ also.

The relation on norms is an equivalence relation so also the opposite is true.

  • $\begingroup$ would u please explain the observation u mentioned? I'm struggling with myself to solve it but i can't understand how u reached it. $\endgroup$ Apr 14, 2017 at 13:34
  • $\begingroup$ I made (wrongly) my observation on base of $||x||\leq M||x||'$. Sorry for the confusion I caused $\endgroup$
    – drhab
    Apr 14, 2017 at 13:54
  • $\begingroup$ So abusively switched accent sign. The answer is repaired now. $\endgroup$
    – drhab
    Apr 14, 2017 at 19:20

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