Question about norms and coarseness of topology 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!

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 $T_A \subset T_B$.
