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In the paper (or refer to this answer to view the full proof of this result):

Tsing N.K. [1984]. Infinite dimensional Banach spaces must have uncountable basis—an elementary proof. Amer. Math. Monthly, 96 (5), 505-506.

I cannot show in the very last line that $\displaystyle\frac{1}{2}t_m r_m$ does not approach to $0$ as $m$ increases, so we cannot guarantee that $\| u_{n} - u \|$ does not approach $0$. In fact $r_{m+1} \le r_m$ since we're taking the infimum of a set that is becoming larger. What am I missing here?

tcmtan
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    It is the index $n$ which increases. $m$ is fixed by the choice of $u$, so $\frac{1}{2}t_mr_m$ is a positive constant. – KeeperOfSecrets Oct 15 '21 at 08:58
  • Ah, yes! I can't believe I missed that. Thanks for pointing this out. I wanted to comment in the original post but it's an old thread so I wasn't sure if I was going to get a reply. – tcmtan Oct 15 '21 at 22:10

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