4
$\begingroup$

I've been thinking about norms and asked myself the following question:

If I have two norms $\|\cdot\|_A$ and $\|\cdot\|_B$ with $\|\cdot\|_A \leq \|\cdot\|_B$, which topology is coarser, that is, has less open sets?

I tried to answer it as follows, is this correct?

It is enough to think about the ball of radius one around zero. Since $\|\cdot\|_A \leq \|\cdot\|_B$, there are more poins in $B_{\|\cdot\|_A}(0,1)$ than in $B_{\|\cdot\|_B}(0,1)$. In particular, there is a point that is in $B_{\|\cdot\|_A}(0,1)$ but not in $B_{\|\cdot\|_B}(0,1)$. Around this point we cannot make an epsilon $A$-ball that is contained in the $B$-unit-ball. Hence the $B$-unit ball is not open in the $A$-topology, hence the $B$-topology is coarser than the $A$-topology.

Thanks for help!

$\endgroup$
  • 2
    $\begingroup$ If, for instance, $\|x\|_A=\frac12\|x\|_B$ for all $x$, the two topologies are the same. Your sentence that begins Around this point is false in general. $\endgroup$ – Brian M. Scott Sep 20 '12 at 9:29
  • $\begingroup$ @BrianM.Scott Right. But if there is no constant such that $c \|\cdot\|_A = \|\cdot\|_B$ then I should be able to find a sequence that converges in the $A$-topology but diverges in the $B$-topology. Right? $\endgroup$ – Rudy the Reindeer Sep 20 '12 at 10:29
  • 1
    $\begingroup$ Nope, you can have equivalent norms that aren't constant multiples of each other, the best-known example being the Manhattan norm and the Euclidean norm in real space. The former has open cubes for basis elements, the latter open balls, but since there's a cube inside every ball and vice versa, convergence in one norm is equivalent to convergence in the other. $\endgroup$ – Kevin Carlson Sep 20 '12 at 12:38
  • $\begingroup$ @BrianM.Scott I drew picture. How can I make $A$-ball around red point that is contained in $B$-unit ball? $\endgroup$ – Rudy the Reindeer Sep 20 '12 at 15:12
  • 1
    $\begingroup$ @bananalyst Of course, I'll upvote it! $\endgroup$ – Norbert Sep 22 '12 at 7:56
1
$\begingroup$

Let $\|\cdot\|_A \leq \|\cdot\|_B$.

Consider the open unit ball $B_A (0,1)$. Let $x \in B_A (0,1)$. Since $B_A (0,1)$ is open with respect to $\|\cdot\|_A$, there exists an $\varepsilon$ such that $B_A(x,\varepsilon) \subset B_A(0,1)$. Since $\|\cdot\|_A \leq \|\cdot\|_B$, $B_B(x,\varepsilon) \subset B_A(x,\varepsilon)$, hence $B_A(0,1)$ is also open in the $B$-topology.

Hence from $\|\cdot\|_A \leq \|\cdot\|_B$ we can conclude that $T_A \subset T_B$.

$\endgroup$

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.