I'm not quite sure about this but I wrote down during a lecture that a normed vector space with an algebraic basis containing a countably infinite number of elements is never complete.

What I mean by algebraic basis is that the elements of this vector space are the finite combinations of those from the basis.

I'm wondering whether the statement should actually contain "infinite" instead of "countably infinite" but I'm not even sure.

Can anyone tell me whether this statement is true or correct me if it's close to being true ?


Yes, it is true: if a normed vector space has a countable Hamel basis, then it cannot be complete. This is intuitively easy to grasp. Let $\{e_n\,|\,n\in\mathbb N\}$ be such a basis. You can assume, withou loss of generality, that each $e_n$ has norm $1$. Consider the series $\sum_{n=1}^\infty\frac1{n^2}e_n$. If the space was complete, then this series would converge. But you cannot express its sum as a finite linear combination of the $e_n$'s.

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  • $\begingroup$ This proof seems to me to work for uncountably infinite bases too. Which would mean that a normed vector space is complete iff it has finite dimension. Is this right ? I've always heard that finite dimension meant complete, but never the converse, so I'm a little surprised. $\endgroup$ – James Well Dec 10 '17 at 11:17
  • $\begingroup$ @JamesWell No, it does not work for an uncountable basis. In my proof, I used the fact that the sum of the series must be a finite linear combination of the $e_n$'s (since they for a basis). Why should this hold if the base was uncountable? The sum of the series might well be, say another element of the basis outside the set $\{e_n\,|\,n\in\mathbb N\}$. $\endgroup$ – José Carlos Santos Dec 10 '17 at 12:07
  • $\begingroup$ I made the mistake of thinking that if $(e_i)_{i\in I}$ were independent then the series could not be equal to a member not involved in the series - but in fact, linear independence just means no finite sum of them can. $\endgroup$ – James Well Dec 10 '17 at 14:04

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