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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.2012-09-20
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    @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?2012-09-20
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    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.2012-09-20
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    @BrianM.Scott I drew [picture](http://i.stack.imgur.com/uVYza.png). How can I make $A$-ball around red point that is contained in $B$-unit ball?2012-09-20
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    (picture drawn using [flockdraw](http://flockdraw.com/)).2012-09-20
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    @BrianM.Scott I think I was confusing things. We want a $B$-ball around the red point contained in the $A$-ball. And an $A$-ball around every point inside the $B$-ball contained in the $A$-unit-ball.2012-09-20
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    @BrianM.Scott So from $\|x\|_A \leq \|x\|_B$ we cannot deduce anything about which topology is coarser. Correct?2012-09-20
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    No, the other way should work: we can deduce that every set open in $A$ is also open in $B$.2012-09-20
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    @bananalyst now you able to answer your question2012-09-20
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    @Norbert Yes! :) Should I post an answer?2012-09-20
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    @bananalyst Of course, I'll upvote it!2012-09-22
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    @Norbert I did it!2012-09-24

1 Answers 1

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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 $T_A \subset T_B$.