Suppose is an n-dimensional normed space over (or ) with basis . There exists a such that

for *any* selection of in the field.

**Proof**: Let and let . Now our original inequality is equivalent to

for all that satisfy . Suppose no such exists. Then we can construct a sequence where , with for each , and . Now note that since implies if we look at the sequence (where is fixed), that we have a bounded sequence, and so we must have a convergent subsequence. Apply this for , let the limit of that subsequence be , and let be the associated subsequence of the original sequence. On that subsequence, apply again for , then on that subsequence again on and so on until we have

Now note that each (where is the limit of the subsequence with the associated from earlier). This implies . Since we required for each earlier, we now have that . This means that , so the subsequence converged to a non-zero element, which means the original sequence can not converge to the zero element. We could only do this because we assumed no such existed, so a with the desired property must exist.

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