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(Lael and Nourouzi 2008) Let $(V, \mu, \nu)$ be an $I F$-normed space. Assume further that $\mu(x, t)>0$ for all $t>0$ implies $x=0$. Define
$\|x\|_\alpha=\inf \{t>0: \mu(x, t)>\alpha \text { and } v(x, t)$
where $\alpha \in(0,1)$. Then $\left\{\|x\|_\alpha: \alpha \in(0,1)\right\}$ is an ascending family of norms on $V$.
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To show that $\{\|x\|_\alpha: \alpha \in (0,1)\}$ is an ascending family of norms, we need to show that for any $0\lt \alpha_1\lt;\alpha_2\lt 1$ and any $x \in V$, we have $\|x\|_{\alpha_1} \leq \|x\|_{\alpha_2}$.Let $x \in V$ be arbitrary. By definition, we have
$\|x\|_{\alpha_1} = \inf\{t \gt 0: \mu(x,t) \gt \alpha_1 \text{ and } \nu(x,t) \leq 1\}$
and
$\|x\|_{\alpha_2} = \inf\{t \gt;0: \mu(x,t) \gt \alpha_2 \text{ and } \nu(x,t) \leq 1\}.$
Since $\alpha_1 \lt \alpha_2$, we have
\[
\{t&gt;0: \mu(x,t) \gt \alpha_1 \text{ and } \nu(x,t) \leq 1
by Diamond (50,343 points)
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