# finer topology and topology induced by norms

I have two questions. First: Let $$d_1,d_2$$ be metrics in $$M$$ and consider the open balls $$B_r^1(x)=\{y \in M; d_1(x,y) $$B_r^2(x)=\{y \in M; d_2(x,y)

a) Proof that, if for all $$x \in M$$ and $$r>0$$ there is $$\delta>0$$ satisfying $$B_\delta^2(x) \subset B_r^1(x)$$, then the topology $$\tau_2$$ generated by $$d_2$$ is finer then $$\tau_1$$ generated by $$d_1$$.

b)Consider two norms $$\|\cdot\|_1,\|\cdot\|_2$$ in a vector space. If we have $$\alpha\|x\|_1\leq\|x\|_2\leq \beta \|x\|_1$$ then $$\|\cdot\|_1,\|\cdot\|_2$$ generate the same topology.

For the first question, i just need to prove that every open ball in $$\tau_1$$ is in $$\tau_2$$? Let $$a \in B_r^1$$, then $$d_1(x,a) and proof that $$d_2(x,a)<\delta$$? I'm just confused with this.

For the second question, i saw this question: Equivalent metrics generate the same topology. We have the same steps? Sounds logical to me, but my textbook does not have nothing on topology induced by norms.

• It doesn’t matter if the metric is generated by a norm or not, only those inequalities matter. Mar 8, 2021 at 6:28

As to a): Let $$O$$ be open for $$d_1$$. Let $$x \in O$$. Then there is some $$r>0$$ such that $$B_r^1(x) \subseteq O$$ as $$x$$ is a $$\tau_1$$-interior point of $$O$$.

Then the condition ensures that there some $$\delta>0$$ such that $$B_\delta^2(x) \subseteq B_r^1(x)$$, so that $$x$$ is a $$\tau_2$$ (or $$d_2-$$) interior point of $$O$$. As $$x \in O$$ was a arbitrary, $$O \in \tau_2$$. So $$\tau_1 \subseteq \tau_2$$ and the claim has been shown.

For the rest, see my answer at the quoted question here. We really apply $$(a)$$ twice one you realise how the condition is fulfilled from the inequalities. The inequalities for the norm immediately imply those for the corresponding metrics $$d(x,y)=\|x-y\|$$ with the same constants, of course. So the fact that we have a norm is not really relevant.

• Thank you! My proof looks a lot like yours. Mar 9, 2021 at 16:00