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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.

Martin Sleziak
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Sushil
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    Any attempts so far? –  Jun 26 '15 at 16:46
  • I've deleted an answer based on a misreading of the question. – Noah Schweber Jun 26 '15 at 17:02
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    A bit of Googling gives this. – David Mitra Jun 26 '15 at 17:07
  • @DavidMitra why the inequality t(n+k) <= (1/3)^n+k-1 hold in the proof given. can you please explain? – Sushil Jun 26 '15 at 17:41
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    @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). – Martin Sleziak Jun 26 '15 at 18:31
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    (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. – Asaf Karagila Jun 26 '15 at 18:34
  • @AsafKaragila Basically question was in ZF with axiom that dimension is well defined. And we need to prove the same as stated in answer: No infinite-dimensional normed linear space with a countable Hamel basis can be complete – Sushil Jun 27 '15 at 15:42
  • Okay. That is an easy consequence of BCT by noting two things: (1) BCT is provable for separable metric spaces without using the axiom of choice; (2) If $V$ has a countable Hamel basis, it is separable. Even if it has other uncountable bases as well. – Asaf Karagila Jun 27 '15 at 15:45
  • Okay this fact I need to prove but beleiving on you. If a vector space has a countable base can it still have uncountable base if I don't assume axiom that dimension is well defined @AsafKaragila – Sushil Jun 27 '15 at 15:57
  • You can find the proof of (1) in Herrlich "The Axiom of Choice", on the part dealing with Baire category theorem; (2) is quite easy to show, simply look at the linear combinations of your Hamel basis with only rational coefficients. To the second question, I don't know the answer, and I recall saying that in a comment to you within the last two days. – Asaf Karagila Jun 27 '15 at 16:01
  • Indeed, within the last two days. – Asaf Karagila Jun 27 '15 at 16:09
  • @AsafKaragila Yes you have said it. I remember. I wanted to ask in the model where it is proved without AC that there exist vector space with Hamel Basis of different cardinalities. Is in that model cardinalities are uncountable(Although it will not answer my question just want to know as fact will prove when I'll study Model Theory) – Sushil Jun 27 '15 at 16:12
  • You can find this in Jech "The Axiom of Choice" as one of the problems in chapter 10, he provides hints and the gist of the proof as given by Lauchli in his original German paper. – Asaf Karagila Jun 27 '15 at 16:14

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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 $X$.

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Martin Sleziak
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  • 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. – Martin Sleziak Jun 26 '15 at 18:28
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    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)? – David Mitra Jun 26 '15 at 18:35
  • 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.) – Martin Sleziak Jun 26 '15 at 18:41
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    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. – Asaf Karagila Jun 26 '15 at 18:46
  • @Martin: Hi Martin. What is $ \theta $ at the start of the second paragraph of the proof? – Transcendental Jun 26 '15 at 23:44
  • @Transcendental it is 0 not θ – Sushil Jun 27 '15 at 02:50
  • @Transcendental The author uses this notation to denote the zero vector. – Martin Sleziak Jun 27 '15 at 03:19
  • @MartinSleziak Can we also prove that Hamel Basis of Banach space will of cardinality atleast c(without using AC or Hahn Banach Theorem). We have to assume CC atleast as mentioned in question: http://math.stackexchange.com/q/1340939/168520 – Sushil Jun 27 '15 at 15:40
  • @MartinSleziak so we would like to prove: No infinite-dimensional normed linear space with a Hamel basis having cardinality strictly less than c can be complete – Sushil Jun 27 '15 at 15:43
  • I don't know @Sushil. You can either update your question, or post a new one (or try to ask in chat, although there is less chance that somebody notices it). – Martin Sleziak Jun 27 '15 at 15:55
  • I see that you have posted a question about it: http://math.stackexchange.com/questions/1341283/banach-space-and-hamel-basis-cardinality – Martin Sleziak Jun 27 '15 at 16:08