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$X$ is a $\mathbb{K}$ normed linear space. To show: \begin{equation} \cup_{1\le p < \infty}\{x\in \mathbb{K}^n:||x||_{p}<1\} = \{x\in \mathbb{K}^n:||x||_{\infty}<1\} \end{equation}


I have already shown that the 1st set is contained in the second. And for the 2nd part I think we haveto use $lim_{p\rightarrow \infty}||x||_p=||x||_\infty$, some how.


My approch: Let the second set not be a subset of the first. If we concider a point in the second set, we can find a open ball around it which doesn't intersect with the first set since both the sets are open. Now using the limiting condition we can show that the distance between two norms can be made as close as possible by inceasing $p$. Hence such a open set is not possible. Is this okay? or it lacks rigor?

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  • $\begingroup$ I would say that your current formulation lacks rigor, but if you vigorously prove that the two norms can be made arbitrarily similar. $\endgroup$
    – Neil
    Sep 20 '16 at 19:38
  • $\begingroup$ Can it be done that way? $\endgroup$
    – mm-crj
    Sep 20 '16 at 19:40
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    $\begingroup$ You'd need to demonstrate the leftmost set is open; the rightmost set is open in the topology defined by the metric defined by the $\infty$-norm, but the latter isn't necessarily so. $\endgroup$
    – Neil
    Sep 20 '16 at 19:44
  • $\begingroup$ Any other approaches are also welcome. $\endgroup$
    – mm-crj
    Sep 20 '16 at 19:52
  • $\begingroup$ I would simply take an arbitrary element of the second set, that is, a bounded sequence where no term is greater than or equal to 1, and prove that it must necessary have a p for which its norm converges to below 1. $\endgroup$
    – Neil
    Sep 20 '16 at 19:54
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Let $x\in \cup_{1\le p < \infty}\{x\in \mathbb{K}^n:||x||_{p}<1\}$. Then there is $p\ge 1$ such that $\|x\|_p<1$. Therefore, for each $i\in\{1,2,\cdots,n\}$, $$|x_i|\le\|x\|_p<1$$ which implies $\|x\|_\infty\le \|x\|_p<1$. So $x\in\{x\in \mathbb{K}^n:||x||_{\infty}<1\}$. On the other hand, let $x\in\{x\in \mathbb{K}^n:||x||_{\infty}<1\}$. Then for $i\in\{1,2,\cdots,n\}$, $|x_i|<1$. Without loss of generality, let $$ |x_1|\le |x_2|\le\cdots\le |x_n|, x_n\neq0. $$ Then $$ \|x\|_p\le n^{\frac1p}|x_n|. $$ Noting $\lim_{p\to\infty}n^{\frac1p}=1$, there is $p>1$ such that $n^{\frac1p}<\frac{1}{|x_n|}$. With this $p>1$, $$ \|x\|_p\le n^{\frac1p}|x_n|<1. $$ Thus $x\in \cup_{1\le p < \infty}\{x\in \mathbb{K}^n:||x||_{p}<1\}$. So $$ \cup_{1\le p < \infty}\{x\in \mathbb{K}^n:||x||_{p}<1\}=\{x\in \mathbb{K}^n:||x||_{\infty}<1\}.$$

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