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

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 $d_1(x,a)<r$ 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.

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

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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.

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  • $\begingroup$ Thank you! My proof looks a lot like yours. $\endgroup$
    – ask 1234
    Mar 9, 2021 at 16:00

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