Continuous embedding and topologies Let $X_1, X_2$ Banach spaces. Suppose that $X_1$ is continuous embedded into $X_2$, i.e. $X_1 \subset X_2$ and the injection map $i: X_1 \to X_2, i(x)=x$ for $x \in  X_1$ is linear and continuous:
$$|i(x)|_{X_2}=|x|_{X_2} \leq M |x|_{X_1}$$
for some $M>0$.
Then the norm on $X_1$ is stronger the norm in $X_2$ for the elements $x \in X_1 \subset X_2$.
Does one says then that the norm on $X_1$ is stronger the norm in $X_2$ in general or that the topology of $X_1$ is stronger than the one on $X_2$ (even if I don't know if you can compare two topologies on two different spaces, even though $X_1 \subset X_2$)?
 A: Let $d, e$ be metrics on $X_1$ where $d(x,y)=\|x-y\|_1$ and $e(x,y)=\|x-y\|_2.$ Let $T_d, T_e$ be the topologies on $X_1$ generated by $d,e$ respectively. If $M\in\Bbb R^+$ and if $e(x,y)\le M\cdot d(x,y)$ for all $x,y\in X_1$ then $T_e\subseteq T_d.$
Example. Let $X_1=l^1$ be the set of real sequences $(x_n)_{n\in\Bbb N}$ such that  $\infty>\sum_{n\in\Bbb N}|x_n|=\|(x_n)_{n\in\Bbb N}\|_1.$ Let $X_2=l^{\infty}$ be the set of real sequences $(x_n)_{n\in\Bbb N}$ such that $\infty>\sup_{n\in\Bbb N}|x_n|=\|(x_n)_{n\in\Bbb N}\|_{\infty}.$
In this case we have $M=1.$
The open $l^1$-ball $B_1(\bar 0,1)=\{x\in l^1:\|x\|_1<1\}$ of radius $1$ centered at the origin $\bar 0=(0,0,0,...)$ is NOT open in the topology on $X_1$ generated by the $l^{\infty}$ norm.
If it were, then for some $r\in\Bbb R^+$ we would have $B_1(\bar 0,1)\supset B_{\infty}(\bar 0,r)= \{x\in l^1: \|x\|_{\infty}<r\}.$
But  take $m\in\Bbb N$ with $m>2/r \,;$ let $x_n=r/2$ if $n\le m$ and let $x_n=0$ if $n>m.$ Then $(x_n)_{n\in\Bbb N}\in B_{\infty}(\bar 0,r)\setminus  B_1(\bar 0,1).$
So the $l^{\infty}$ norm on the $set$ $l^1$ generates a strictly weaker topology than the $l^1$ norm.
