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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!

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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. – Brian M. Scott Sep 20 '12 at 9:29
@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? – Rudy the Reindeer Sep 20 '12 at 10:29
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. – Kevin Carlson Sep 20 '12 at 12:38
@BrianM.Scott I drew picture. How can I make $A$-ball around red point that is contained in $B$-unit ball? – Rudy the Reindeer Sep 20 '12 at 15:12
@bananalyst Of course, I'll upvote it! – Norbert Sep 22 '12 at 7:56
up vote 1 down vote accepted

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 $T_A \subset T_B$.

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