Prove that every Hamel basis in an infinite dimensional Banach Space is uncountable without using Baire Category Theory. We are assuming axiom that vector space dimension (if exist) is well-defined.

Note: Axiom that vector space dimension(if exist) is well-defined is independent of DC. See Sizes of bases of vector spaces without the axiom of choice at MathOverflow.

  • 2
    $\begingroup$ Any attempts so far? $\endgroup$ – user228113 Jun 26 '15 at 16:46
  • $\begingroup$ I've deleted an answer based on a misreading of the question. $\endgroup$ – Noah Schweber Jun 26 '15 at 17:02
  • 3
    $\begingroup$ A bit of Googling gives this. $\endgroup$ – David Mitra Jun 26 '15 at 17:07
  • 1
    $\begingroup$ @Sushil For $k=2$ you have $t_{n+2}=\frac13r_{n+1}t_{n+1} \le \frac13t_{n+1} \le \frac1{3^2}r_nt_n$. You can continue similarly for bigger $k$'s (by induction). $\endgroup$ – Martin Sleziak Jun 26 '15 at 18:31
  • 1
    $\begingroup$ (1) Is this a question in $\sf ZF$ or in $\sf ZFC$? (2) $\sf DC$ has absolutely nothing to do with "dimension is well-defined", since it is an axiom which has to do with the family of "consequences of BPI". It is not "weaker" than $\sf DC$, because I very much doubt that it is at all provable from $\sf DC$. In this context you could be asking if $\sf ZF$ proves that the dimension of a Banach space is well-defined (namely, if a Hamel basis exists, then all Hamel bases have the same cardinality), again I don't know the answer, and I doubt this appears in the literature directly. $\endgroup$ – Asaf Karagila Jun 26 '15 at 18:34

As David Mitra mentioned in his comment, one such proof can be found in Morrison T.J. Functional Analysis. An Introduction to Banach Space Theory (Wiley, 2000), p.221.

The proof we give is very elementary and avoids the usual argument seen for this fact, which involves either the Baire Category Theorem or the Hahn-Banach Theorem. It is due to the Chinese mathematician Nam-Kiu Tsing (1984).

Proposition 5.1. No infinite-dimensional normed linear space with a countable Hamel basis can be complete.

Proof. Let $(X,\|\cdot\|)$ be a normed linear space with Hamel basis $(e_n)_n$, and note that without loss of generality we can assume $\|e_n\|=1$. Let $S_{n-1}$ denote the linear subspace of $X$ spanned by $\{e_1,e_2,\dots,e_{n-1}\}$, and let $r_n \equiv \inf\{ \|x+e_n\|; x\in S_{n-1}\}$ for any $n\ge2$.

Since $\theta\in S_{n-1}$, it follows that $r_n\le \|0+e_n\| =1$ for all $n\ge2$. Now since $S_{n-1}$ is finite-dimensional, it is complete, and hence closed in $X$. Since $e_n\notin S_{n-1}$, we have that $r_n>0$ for all $n\ge2$. Now define the following scalar sequence $(t_n)_n$ by $t_1=1$, $t_2=\frac13$ and for $n\ge2$, $t_{n+1}=\frac13r_nt_n$. Then note that we have $$0<t_{n+k}\le\left(\frac13\right)^k r_nt_n \le \left(\frac13\right)^{n+k-1}$$ for all $n\ge2$ and $k\in\mathbb N$. Now for each $n\in\mathbb N$, define $u_n=\sum_{i=1}^n t_i e_i$ and note that $(u_n)_n$ is a cauchy sequence in $X$. But also notice that for any element $u=\sum_{i=1}^{m-1} \alpha_i e_i\in X$, we have \begin{align} \| u_{n} - u \| & = \left\| \sum_{i = 1}^{m - 1} (t_{i} - \alpha_{i}) e_{i} + t_{m} e_{m} + \sum_{i = m + 1}^{n} t_{i} e_{i} \right\| \\ & \ge \left\| \sum_{i = 1}^{m - 1} (t_{i} - \alpha_{i}) e_{i} + t_{m} e_{m} \right\| - \left\| \sum_{i = m + 1}^{n} t_{i} e_{i} \right\| \\ & \ge t_{m} \left\| \sum_{i = 1}^{m - 1} \frac{1}{t_{m}} (t_{i} - \alpha_{i}) e_{i} + e_{m} \right\| - \sum_{i = m + 1}^{n} t_{i}\\ & \ge t_{m} r_{m} - \sum_{i = 1}^{n - m} \left( \frac{1}{3} \right)^{i} r_{m} t_{m} \\ & \ge \frac{1}{2} t_{m} r_{m} \end{align} for all $n>m$. But this means that $\|u_n-u\|$ does not go to zero, which implies $(u_n)_n$ does not converge. Hence, $X$ is not complete.

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

The symbol $\theta$ is used to denote the zero vector of the space $X$.

  • $\begingroup$ Maybe I should add that this book does not give any analysis whether AC is used at some place of the proof. The OP seems to be after a proof in ZF. $\endgroup$ – Martin Sleziak Jun 26 '15 at 18:28
  • 1
    $\begingroup$ I may be off, but I think the exponent of "$1/3$" in the second to last inequality should be "$i$" (typo in the book, I suppose)? $\endgroup$ – David Mitra Jun 26 '15 at 18:35
  • $\begingroup$ I have corrected it. The book indeed writes $\left(\frac13\right)^m$, but in this way it would work. (I did not search whether there is a newer edition of this book or whether errata are published somewhere.) And it is written the same way as you suggest in Tsing's paper. (Modulo somewhat different notation.) $\endgroup$ – Martin Sleziak Jun 26 '15 at 18:41
  • 1
    $\begingroup$ BCT for separable metric spaces is true in $\sf ZF$. And this here is more or less a proof for that fact (needs a bit of massaging, I guess). So really just noting that a countable Hamel basis implies separability, and applying BCT is a proof "without BCT". Not to mention that the Hahn-Banach theorem is not provable from $\sf DC$ (nor it implies it), so proving this from HB would also be considered a valid answer. $\endgroup$ – Asaf Karagila Jun 26 '15 at 18:46
  • $\begingroup$ @Martin: Hi Martin. What is $ \theta $ at the start of the second paragraph of the proof? $\endgroup$ – Transcendental Jun 26 '15 at 23:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.