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The following is from Functional Analysis book by Conway:

Let $X$ be a separable infinite-dimensional Banach space and let ${\{e_i : i \in I}\}$ be a Hamel basis for $X$ with $|| e_i || = 1$ for all $i$. Note that a Baire Category argument shows that I is uncountable.

I am confused with two things:

1- How Baire Category Thm is related?

2- I think the book is wrong because I must be countable not uncountable? See this, for example.

  • It's a way to show that if $X$ is an infinite dimensional Banach space then its Hamel basis is uncountable. Also you see the term basis as in a Schauder basis, but that is not a Hamel basis. – Anonmath101 Jun 28 '21 at 12:35
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    For 2, the link refers to orthonormal bases (which admit representations as infinite sums), Conway is using Hamel bases, which use only finite sums. – David Mitra Jun 28 '21 at 12:37
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    There is a difference between Hamel and Hilbert bases. Every separable Hilbert space admits a countable Hilbert basis, but every infinite-dimensional vector space can only admit an uncountable Hamel basis. – amnesiac Jun 28 '21 at 12:38
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    There is a difference between a Hamel basis and an orthonormal basis. No Banach space can have a countable Hamel basis. – Kavi Rama Murthy Jun 28 '21 at 12:38
  • @DavidMitra the problem still remains : with uncountable basis there can't be a dense countable set (?) –  Jun 28 '21 at 12:41
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    For 1, see this. – David Mitra Jun 28 '21 at 12:42
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    @amnesiac "every infinite-dimensional vector space can only admit an uncountable Hamel basis" is only true if you restrict to complete spaces. Otherwise a counterexample is the space of finite sequences of reals. – Robin Saunders Jun 28 '21 at 21:16
  • Similarly, @KaviRamaMurthy Banach spaces can be finite-dimensional! – Robin Saunders Jun 28 '21 at 21:18
  • @RobinSaunders By countable Hamel basis I meant countably infinite Hamel basis. – Kavi Rama Murthy Jun 28 '21 at 22:11
  • Ah, fair enough. I think this is a fairly common ambiguity, maybe the convention varies by area. – Robin Saunders Jun 28 '21 at 22:44

2 Answers2

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Let $X$ be a Banach space. Then, either $X$ has a finite Hamel basis (and in this case $X$ is finite-dimensional) or $X$ has an uncountable Hamel basis.

Proof: First, note that using Zorn lemma, we can prove that every vector space has a Hamel basis. (If you need, I can add details here).

Let $X$ be a Banach space. To get a contradiction, let us suppose that $X$ has a countable infinite Hamel basis $B = \{ e_i \}_{i \in \Bbb N}$.

For each $n \in \Bbb N$, let $V_n$ be the subspace generated by $\{e_0, \cdots, e_n\}$. It is clear that $V_n$ is a finite dimensional subspace of $X$. So $V_n$ is closed.

Since, for all $n \in \Bbb N$, $V_n$ is a proper subspace of $X$, it is also clear that $V_n$ has empty interior in $X$ (see Remark 2).

Now, because $B$ is a Hamel basis, we have that $X= \bigcup_{n \in \Bbb N} V_n$.

However, by Baire Category Theorem, a complete metric space (in our case $X$) can not be the countable unions of closed subsets having empty interior (nowhere dense closed subsets). Contradiction. So $X$ can not have a countable infinite Hamel basis.

Remark 1:

  1. The fact that, for all $n \in \Bbb N$, $V_n$ is closed and has empty interior remains true even if $X$ is just an infinite-dimensional normed vector space.

  2. $X= \bigcup_{n \in \Bbb N} V_n$ is true because $B$ is a Hamel basis

  3. It is only to get the contradiction to Baire Category Theorem, that we need $X$ to be a Banach space (so $X$ is a complete metric space).

Remark 2: Let $X$ be any normed vector space and $V$ be any proper subspace of $X$. Then $V$ has empty interior in $X$.

Proof: Let us prove the counter-positive. Suppose $V$ has an interior point $p$. It means that there is $\varepsilon >0$, such that $B_X(p, \varepsilon) \subseteq V$. Since $p \in V$ and $V$ is a vector space, we have that $$ B_X(0, \varepsilon)= \{y - p : y \in B_X(p, \varepsilon) \} \subseteq V - p =V$$

Now, given any $x \in X$, let $K > \frac{ \|x\|}{\varepsilon}$. So $ \| K^{-1} x \| < \varepsilon $. So $ K^{-1} x \in V$. Since $V$ is a vector space, $x = K(K^{-1} x) \in V$. So $X \subseteq V$. It means that $V$ is NOT a proper subspace of $X$.

So we have that, if $V$ is a proper subspace of $X$, then $V$ has empty interior in $X$.

Ramiro
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  • I couldn't understand two things : 1-how "$V_n$ has empty interior in $X$" holds? 2-why "Chris Culter" argument in OP's link fails?(i.e. I can make that argument again because X is separable and make the basis countable!) –  Jun 28 '21 at 19:35
  • I had learned about the first paragraph from Ex12:15.1a Bruckner's RA :) –  Jun 28 '21 at 20:00
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    @L.G. For the "empty interior", I added a detailed proof as Remark 2. Please, take a look. – Ramiro Jun 28 '21 at 20:02
  • I want to figure out how $B_X(0, \varepsilon)= {y - p : y \in B_X(p, \varepsilon) } \subseteq V -p$ can be turned to $B_X(0, \varepsilon)= {y - p : y \in B_X(p, \varepsilon) } \subseteq V$. The answer lies in the sentence just before that, which I am trying to understand –  Jun 28 '21 at 20:13
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    @L.G. $V$ is a vector space, $p \in V$, so $V - p = V$. – Ramiro Jun 28 '21 at 20:19
  • Sorry Ramiro, I still am confused why exactly the argument of Chris Culter in https://math.stackexchange.com/questions/450106/uncountable-basis-and-separability fails here? Same argument is repeated in Bruckner's and also Kreyszig's book for separable sapces –  Jun 28 '21 at 20:23
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    @L.G. It does not fail. It can combined with the above result and then we have: Given any infinite-dimensional separable Hilbert space $X$, then any Hamel basis for $X$ will not be ortonormal. – Ramiro Jun 28 '21 at 20:33
  • Thank you very much Ramiro :) –  Jun 28 '21 at 20:45
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There is a dichotomy of the cardinality of Hamel bases. A Hamel basis of a Banach Space is either finite (then the Banach Space is finite dimensional) or it is uncountable (if the space is of infinite dimension) a prove of this using Baire's Category Theorem can be found here:

https://math.stackexchange.com/a/217843/860249

To get the notation straight: A Hamel basis is a set $E \subseteq X$ of a $\mathbb{K}$-Banach space $X$ which is linearly independent, i.e., for any distinct $x_1, \dots, x_n \in E$ and coefficients $c_1, \dots, c_n \in \mathbb{K}$ we have

$$ \sum_{k = 1}^n c_i x_i = 0 \implies c_1 = \cdots = c_n = 0, $$

and a generating set, i.e., for all $x \in X$ there exist $x_1, \dots, x_n \in E$ and coefficients $c_1, \dots, c_n \in \mathbb{K}$ such that

$$ x = \sum_{k = 1}^n c_i x_i.$$

Notice that the sum here is always finite. With Schauder or Hilbert bases this need not be the case. Intuitively this requires that the Hamel bases to be larger, as less elements $x \in X$ can be represented using finite sums than using infinite series.

Julian Hölz
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