Do there exist in infinitely dimensional normed spaces linearly independent and dense subsets? (Existence of linearly independent dense subset is equivalent of existence of dense Hamel Basis.)


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    $\begingroup$ What do you mean by linear independent subset? $\endgroup$ – Davide Giraudo Aug 27 '11 at 9:56
  • $\begingroup$ @Davide Giraudo: I'd assume he means a subset which is linearly independent and dense. $\endgroup$ – Ilmari Karonen Aug 27 '11 at 10:39
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    $\begingroup$ In the discussion leading up to Problem 9 in Halmos's Hilbert space problem book, this is answered for separable Hilbert space, using the same method as in Nate Eldredge's answer. $\endgroup$ – Jonas Meyer Aug 27 '11 at 16:44

If your space (call it $X$) is separable then the answer is yes.

Pick a countable base $\{U_n\}$ for $X$. Construct the sequence $\{x_1, x_2, \dots\}$ inductively as follows. First choose $x_1 \in U_1$ arbitrarily. Now for the inductive step, if $\{x_1, \dots, x_n\}$ are already chosen, set $E_n$ to be their linear span. Since $E_n$ is finite dimensional, it is a proper subspace of $X$, and so it must have empty interior. In particular it does not contain $U_{n+1}$, so choose $x_{n+1} \in U_{n+1} \backslash E_n$.

The sequence $\{x_1, x_2, \dots\}$ is linearly independent by construction, and dense because it intersects every $U_n$.

Note that we only used dependent choice, which is much less than is needed to produce a Hamel basis.

The non-separable case seems more difficult.

Edit: After reading Martin Sleziak's answer I think I see how to handle the non-separable case, at the expense of more choice.

Fix a Hamel basis $\{e_i\}_{i \in I}$ for $X$, where $|I| = \kappa = \dim X$. Let $A$ be the $\mathbb{Q}$-linear span of $\{e_i\}$; then $|A| = \kappa$ also. $A$ is clearly dense in $X$, so $\mathcal{U} = \{B(x,r) : x \in A, r \in \mathbb{Q}\}$ is a base for $X$, and $|\mathcal{U}| = \kappa$ also.

Let $\omega$ be the smallest ordinal of cardinality $\kappa$. We can then enumerate $\mathcal{U}$ as $\mathcal{U} = \{U_j\}_{j < \omega}$. We now produce the desired set $\{x_j\}_{j < \omega}$ by transfinite induction: for $j < \omega$, let $E_j$ be the ($\mathbb{R}$-)linear span of $\{x_m : m < j\}$. Then $\dim E_j = |j| < \kappa$, and in particular $E_j$ is a proper subspace of $X$. As such it has empty interior, so does not contain $U_j$, and so we can pick $x_j \in U_j \backslash E_j$. As before, the set $\{x_j\}_{j < \omega}$ is linearly independent by construction, and is dense because it intersects every $U_j$.

(Sorry if this is not written too clearly; I'm not really used to arguments like this.)

As in Martin's references, the key ingredient is a base of cardinality at most $\dim X$. Of course there can be bases of smaller cardinality, as in the separable case, and you could use such a base to construct a smaller linearly independent dense set.

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  • $\begingroup$ It is great answer. $\endgroup$ – Richard Aug 30 '11 at 8:56

What follows is a LaTeX'ed version of an essay I posted in sci.math on 18 August 2006 that may be of more use here than buried in the sci.math archive (where I suspect few people now bother to search for things like this essay).

Google's sci.math archive of the essay

Math Forum's sci.math archive of the essay


In what follows, many of the results were stated and proved in more general settings, especially the papers below that were published in the late 1940s and later. However, in the interest of simplicity, I'll usually stay within the realm of Banach spaces $X$. The only topology and metric on $X$ that will arise in the statements I give will be those that are derivable from the norm on $X$.

The term "linear subspace" will mean a subset of $X$ in the vector space sense, and the term "hyperplane" will mean a translate of a linear subspace that has co-dimension $1$. Hyperplanes in normed spaces can be characterized as the inverse images of singletons under linear functionals. That is, if $f:X \rightarrow \mathbb R$ is linear and $r$ is a real number, then $f^{-1}[\{r\}]$ is a hyperplane, and every hyperplane in $X$ arises in this way for some linear functional $f$ and real number $r$. If $X$ is a finite-dimensional normed space, then every linear functional is continuous, and hence for finite-dimensional normed (= Banach, in this case) spaces, every hyperplane is closed. In fact, every proper linear subspace in a finite-dimensional normed space is both closed and, in a rather strong way, nowhere dense. Moreover, every proper closed linear subspace in a normed space is nowhere dense (in a rather strong way).

However, for each infinite-dimensional normed space (neither completeness nor separability is needed for this), it is known that discontinuous linear functionals exist. While this doesn't immediately imply that there exist non-closed hyperplanes, at least if we only consider the situation from a topological standpoint (a function will be continuous if the inverse image of each closed set is closed but, in general, not if we only know that the inverse image of each singleton set is closed, as any one-to-one discontinuous function shows), it is known that for each discontinuous linear functional $f$, every hyperplane arising from $f$ is dense in $X$. Thus, every hyperplane in $X$ is either closed in $X$ or dense in $X$. The first two web pages below give a proof of this result, and other proofs of this result (not necessarily different from a mathematical standpoint) can be found in the google-book search results below.





Another result worth mentioning is that every Borel linear subspace in a Banach space (not necessarily separable) is either first category (i.e. meager) in $X$ or equal to $X$. This is proved in Goffman/Pedrick [6] (p. 80).


Let $X$ be a normed space and $V$ be a proper linear subspace of $X$.

$V$ closed in $X$ $\Rightarrow$ $V$ is nowhere dense in $X$.

($X$ Banach) $V$ Borel in $X$ $\Rightarrow$ $V$ is meager in $X$.

I don't know the historical details of any of the above results, but I would imagine that all of them (except possibly the last one) have been known since the late 1920s.

The answers to either of the questions below may be known, may be easy, may be both known and easy, or may be neither known nor easy. I haven't put forth much effort towards trying to answer either question.

QUESTION 1: Can (must?) a proper Borel linear subspace of some (all?) Banach space(s) be meager but not $\sigma$-porous?

NOTE: Proper closed linear subspaces of normed spaces are porous in a rather strong way. To show nowhere dense, simply note that if the closed set $V$ were not nowhere dense, then $V$ would contain an open ball of center $x$ and radius $r>0$. From this it follows that $V$ would contain open balls of center $x$ and radius $kx$ for every $k > 0$ (because $V$ is a linear subspace), which would then contradict $V$ being a proper linear subspace. To show the stronger result that $V$ is porous, make use of F. Riesz's lemma involving proper closed linear subspaces, e.g. Goffman/Pedrick [6] (p. 82). (For why I said "porous in a rather strong way", see the paper cited in Zbl 1101.28300.)

QUESTION 2: Can a proper non-Borel linear subspace of some (all?) Banach space(s) be meager, or even $\sigma$-porous?

NOTE: There exist proper non-Borel linear subspaces that are not meager, e.g. Hausdorff [7].



Let $X$ be a separable Banach space and $V$ be a proper linear subspace of $X$.

$V$ is $G_{\delta}$ $\Rightarrow$ $V$ is closed.

For each countable ordinal $\alpha \geq 1$, $V$ can belong to the additive Borel class $\alpha$ and not to the multiplicative Borel class $\alpha$ (and hence, not to any lower Borel class).

For each countable ordinal $\alpha \geq 2$, $V$ can belong to the multiplicative Borel class $\alpha$ and not to the additive Borel class $\alpha$ (and hence, not to any lower Borel class).

For each countable ordinal $\alpha \geq 2$, $V$ can belong to the ambiguous Borel class $\alpha$ and not to any lower Borel class.

For each positive integer $n$, $V$ can belong to the $n$th projective class and not to any lower projective class.

It is possible for $V$ to not belong to any finite projective class.

The dates below are the received (or other) dates given in the paper.

3 September 1931 -- Hausdorff [7] proved that in every infinite-dimensional Banach space, there exists a linear subspace that isn't $G_{\delta}$. Hausdorff's example was also not first category, and so by the result I cited from Goffman/Pedrick's text above, Hausdorff's example was actually not a Borel set. However, I don't know if Hausdorff says this in his paper, or even if he was aware of the result I cited from Goffman/Pedrick's text. Pettis [14] uses Hausdorff's construction, which relied on a Hamel basis, to construct pathological subgroups of ${\mathbb R}^{n}$.

28 April 1933 -- Mazur/Sternbach [13] proved that in any infinite-dimensional Banach space, each $G_{\delta}$ linear subspace is closed and there exist $F_{\sigma \delta}$ linear subspaces that aren't $F_{\sigma}$. They also asked if every infinite-dimensional Banach space has linear subspaces of arbitrarily high Borel order. According to Klee [10] (p. 189, footnote 2), Banach announced in 1940 that this is true (Banach died in 1945) and Mazur presented a proof in a 1957 conference in Zakopane. A proof is given in Klee [10].

21 June 1933 -- Banach/Mazur [3], improving on one of the results in Mazur/Sternbach [13], showed that in each infinite-dimensional Banach space, there exist $F_{\sigma \delta}$ linear subspaces that aren't $G_{\delta \sigma}$.

22 June 1933 -- Banach/Kuratowski [2] proved that, in any separable infinite-dimensional Banach space, there exist linear subspaces that are $CA$, $CPCA$, $CPCPCA$, etc. but not $A$, $PCA$, $PCPCA$, etc., respectively (these are the projective set classifications where $A$ is analytic, $C$ is complement, and $P$ is projection), as well the existence of a linear subspace that does not belong to any of these projective classes. Their proof involves showing that there exists a co-analytic linear subspace in $C[0,1]$ (sup norm) having certain properties that is not Borel, observing that the method works the same way for the higher level co-projective classes, and then making use of the fact (proved by Banach and Mazur at about the same time) that every separable Banach space can be isometrically embedded into $C[0,1]$. As they were not able to resolve the problem of whether every (or even some, although I suspect universal embedding results would make "some" equivalent to "all") infinite-dimensional Banach space contains an analytic linear subspace that isn't Borel (or any of the corresponding higher level projective versions), they leave this as an open question. Klee answered this in the affirmative, with a proof, in [10].

8 December 1958 -- Klee [10] proved that in any separable infinite-dimensional Banach space we have: (1) For each countable ordinal $\alpha \geq 1$, there exist dense linear subspaces that belong to the additive Borel class alpha and not to the multiplicative Borel class $\alpha$. (2) For each positive integer $n,$ there exist dense linear subspaces that belong to the $n$th projective class and not to any lower projective class.

26 February 1979 -- Mauldin [12] proved that in any separable infinite-dimensional Banach space we have: (1) For each countable ordinal $\alpha \geq 1$, there exist dense linear subspaces that belong to the additive Borel class $\alpha$ and not to the multiplicative Borel class $\alpha$. (2) For each countable ordinal $\alpha \geq 2$, there exist dense linear subspaces that belong to the multiplicative Borel class $\alpha$ and not to the additive Borel class $\alpha$. (3) For each countable ordinal $\alpha \geq 2$, there exist dense linear subspaces that belong to the ambiguous Borel class $\alpha$ (i.e. belongs to both the additive Borel class $\alpha$ and to the multiplicative Borel class $\alpha$) and not to any lower Borel class.

See Klee [11] for an interesting construction of a continuum-length chain of exotic linear subspaces in the Hilbert space $L^{2}[-1,1]$. For more recent miscellaneous results, see Ding/Gao [4] and Farah/Solecki [5].



Let $X$ be a normed space.

For each nonzero cardinal number $b \leq$ card$(X)$, $X$ can be written as a pairwise disjoint union of $b$ many convex sets, each of which is dense in $X$.

In the previous statement, we can strengthen "dense in $X$" to "linearly dense in $X$". (See Klee [9] below.)

The dates below are the received (or other) dates given in the paper.

10 December 1940 -- Turkey [15] (Section 5, p. 101) proved that each infinite dimensional normed space $X$ is a disjoint union of two convex sets, each of which is dense in $X$.

12 October 1948 -- Klee [8]. Let X be an infinite-dimensional Banach space and let $b$ be any nonzero cardinal number less than or equal to card$(X)$. Then $X$ is a pairwise disjoint union of $b$ many convex sets, each of which is dense in $X$.

early 1950? -- Klee [9] improved on the previous result by strengthening "dense" to "ubiquitous". Klee's term "$E$ is ubiquitous in $X$" means that the linear closure of $E$ is equal to $X$, where the linear closure of a set $E$ is defined to be $E \cup \{x \in X:$ there exists $e$ in $E$ such that the half-open segment $[e,x)$ is a subset of $E\}$. [The linear closure of a convex set in ${\mathbb R}^{n}$ is equal to the closure of that convex set.] Incidentally, Klee's result makes sense, and holds, in any infinite-dimensional vector space of the real numbers, and it was in this setting that Klee proved it. Klee mentions at the end of his paper that the sets he used to prove the results in Klee [8] are not ubiquitous.

14 July 1951 -- Pettis [14] observed (p. 614, Theorem 4) that each infinite-dimensional Banach space $X$ is a pairwise disjoint union of continuum many hyperplanes, each of which is dense in $X$. Pettis obtained this result by considering the sets $f^{-1}[\{r\}]$, as $r$ varies over the real numbers, for a discontinuous linear functional $f$.


[1] Stefan Banach, "Théorie des Opérations Linéaires", Monografie Matematyczne #1, 1932, viii + 252 pages.


Search in this book = "Sternbach", choose p. 235, see bottom half of p. 235.

[2] Stefan Banach and Kazimierz [Casimir] Kuratowski, "Sur la structure des ensembles linéaires", Studia Mathematica 4 (1933), 95-99.


[3] Stefan Banach and Stanislaw Mazur, "Eine Bemerkung über die Konvergenzmengen von Folgen linearer Operationen", Studia Mathematica 4 (1933), 90-94.


[4] Longyun Ding and Su Gao, "On separable Banach subspaces", preprint, 28 April 2006, 6 pages.


[5] Ilijas Farah and Slawomir Solecki, "Borel subgroups of Polish groups", Advances in Mathematics 199 #2 (30 January 2006), 499-541.

See p. 503 for some brief historical notes, or see p. 5 (1st URL) or p. 6 (2nd URL):



[6] Casper Goffman and George Pedrick, "First Course in Functional Analysis", 2nd edition, Chelsea Publishing Company, 1983, 284 pages.


Search in this book = "Blumberg", choose p. 80.

[7] Felix Hausdorff, "Zur Theorie der linearen metrischen Räume", Journal für die Reine und Angewandte Mathematik 167 (1932), 294-311.


[8] Victor L. Klee, "Dense convex sets", Duke Mathematical Journal 16 #2 (June 1949), 351-354.

[9] Victor L. Klee, "Decomposition of an infinite-dimensional linear system into ubiquitous convex sets", American Mathematical Monthly 57 #8 (October 1950), 540-541.

[10] Victor L. Klee, "On the Borelian and projective types of linear subspaces", Mathematica Scandinavica 6 (1958), 189-199.


[11] Victor L. Klee, "Solution to Monthly Problem #5717", American Mathematical Monthly 78 #3 (March 1971), 308-309.

Proposed by W. H. Ruckle: "Without using the axiom of choice (i.e. no Hamel basis) construct a continuum $\{X_{r}: 0 < r \leq 1\}$ of linear subspaces of a Hilbert space $H$ which has the following properties: (a) $X_r$ is dense in $H$ for each $r$; (b) if $r < s$, $X_r$ [is a proper subset of] $X_s$ and $X_{s}/X_{r}$ has uncountable dimension; (c) [the union of all the $X_r$s is not equal to] $H$."

Klee's solution: "Represent Hilbert space as $L^{2}[-1,1]$. For each $r$ in $[0,1]$, let $X_r$ consist of all functions $f$ in $L^{2}[-1,1]$ such that for some $a < r$, the restriction of $f$ to $[a,1]$ is equivalent to a function which assumes only finitely many values. The set $\{X_{r}\}$ fulfills all conditions of the problem."

[12] R. Daniel Mauldin, "On the Borel subspaces of algebraic structures", Indiana University Mathematics Journal 29 #2 (1980), 261-265.

[13] Stanislaw Mazur and Leonard Paul Sternbach, "Über die Borelschen Typen von linearen Mengen", Studia Mathematica 4 (1933), 48-53.


[14] Billy James Pettis, "On a vector space construction by Hausdorff", Proceedings of the American Mathematical Society 8 (1957), 611-616.


[15] John Wilder Turkey, "Some notes on the separation of convex sets", Portugaliae Mathematica 3 (1942), 95-102.

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    $\begingroup$ If anyone was wondering, I realize that what I posted doesn't directly address the question asked. However, I didn't realize this until I had invested a fair amount of time into LaTeX'ing my sci.math post. (I misread, or perhaps more accurately I misremembered the question, thinking it was about dense proper subspaces.) I then rationalized to myself that there still might be some interest, since the results are not very well known and they show in a very dramatic way how different infinite dimensional normed spaces can be from finite dimensional normed spaces. $\endgroup$ – Dave L. Renfro Aug 29 '11 at 22:29
  • $\begingroup$ Thanks. I agree with you that such result are very interesting but not very well known. $\endgroup$ – Richard Aug 30 '11 at 13:09
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    $\begingroup$ Your assertion "$V$ is closed in $X$ or $V$ is dense in $X$" looks wrong to me. For example take $X$ to be a Hilbert space, $Y$ a proper closed subspace, and $V$ dense in $Y$. Or, with $V$ dense in $X$, consider $V\oplus0\subset X\oplus X$. $\endgroup$ – Martin Argerami Sep 7 '12 at 9:05
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    $\begingroup$ @Martin Argerami: I'll look this up when I get home this evening (where all my math things are at) and report back on Monday. However, right now I don't see anything wrong with your observation. I suspect that I'll have to assume that $V$ is a hyperplane rather than a proper linear subspace. Either that, or perhaps change the statement to something like "$V$ is closed in $X$ or $V$ is dense in an infinite-dimensional linear subspace of $V.$ $\endgroup$ – Dave L. Renfro Sep 7 '12 at 18:18
  • $\begingroup$ @Martin Argerami: I've made some minor adjustments in the location that your observation pertains to. I decided to omit the incorrect statement, since a correct formulation was given earlier and any corrected version I can think of tends to distract from the main issues at that location. Also, I've changed the "title" there, I've inserted a missing $V$ that I happened to notice, I've removed an explict google sci.math URL-link, and I've included a Math Forum link for my sci.math post (since I think you now need to be google-registered to see google-archived sci.math posts). $\endgroup$ – Dave L. Renfro Sep 10 '12 at 18:01

From the paper R. R. Phelps: Subreflexive normed linear spaces, Archiv der Mathematik, Volume 8, Number 6, 444-450

Theorem 3.1 (KLEE). Suppose the topological linear space $E$ has a neighborhood basis $\mathcal U$ at the origin such that $\operatorname{card} \mathcal U$ is less than or equal to the dimension of $E$. Then $E$ admits a dense Hamel basis.

A consequence of the above theorem is that every infinite dimensional metric linear space contains a dense Hamel basis. This result is given by MACKEY [8, p. 185] for $\aleph_0$-dimensional normed linear spaces.

G. W. Mackey, On infinite dimensional linear spaces. Trans. Amer. math. Soc. 57, 155--207 (1945).

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I think I can sketch a proof that $\mathbb R$, viewed as a vector space over $\mathbb Q$, has a dense Hamel basis. (Of course, this proof relies heavily on the Axiom of Choice; without it, $\mathbb R$ over $\mathbb Q$ might not have a Hamel basis at all.)

First, pick any Hamel basis $\mathcal H$ for $\mathbb R$ over $\mathbb Q$. Let $f$ be an enumeration of $\mathbb Q$ (i.e. a bijection from $\mathbb N$ to $\mathbb Q$), and let $g$ be an injection from $\mathbb N$ to $\mathcal H$.

Now, for each $n \in \mathbb N$, choose an element $x_n$ from $g(n)\mathbb Q \cap [f(n)-\frac 1n, f(n)+\frac 1n]$. The set $\mathcal X = \{x_1, x_2, \ldots\}$ is obviously linearly independent (over $\mathbb Q$). I claim it is also dense in $\mathbb R$.

To prove this, we need to show that every open interval $(a,b) \subset \mathbb R$, $a < b$, contains some $x_n \in \mathcal X$. Let $\delta = (b-a)/3$, and let $m = 1/\delta$. Since the interval $(a+\delta, b-\delta)$ contains infinitely many elements of $\mathbb Q$, it must contain some $q \in \mathbb Q$ such that $q = f(n)$ for some $n > m$. Thus, $|x_n - q| \le \frac 1n < \frac 1m = \delta$, and so $x_n \in (a,b)$.

Of course, we can also extend $\mathcal X$ to a full Hamel basis by adding $\mathcal H \setminus g(\mathbb N)$ to it.

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  • $\begingroup$ Basically the same approach is given in Komjath, Totik: Problems and theorems in classical set theory, page 317 $\endgroup$ – Martin Sleziak Aug 27 '11 at 14:50
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    $\begingroup$ To overcome the issue with the axiom of choice note the following: the algebraic numbers in $\Bbb R$ are countable and dense; and make a normed space over $\Bbb Q$ as well. Do your construction there and the rest is still linearly independent and dense in $\Bbb R$. $\endgroup$ – Asaf Karagila Nov 6 '13 at 16:45

This is quite an interesting question that I will attempt to answer even though I am not an expert in this field.

The first my thought was to say something about Fourier series or Stone-Weierstrass theorem, but then I realized the the question is whether such subsets exist in all normed spaces.

After a second thought I recalled that these sets have a name Schauder bases. As it is well known, that Hamel bases exist in any vector space. The difference between these two types of bases is that a Hamel basis requires a finite linear combination to represent any element of the vector space, while with a Schauder basis one can take infinite sums provided there is a notion of convergence (as in normed spaces).

Then I did a little googling and I found out that the answer is NO.

According to this "Per Enflo found a separable Banach space that doesn't have a Schauder basis." (Per Enflo is a very interesting mathematician, by the way).

Later I came across another link where this topic was covered in a higher detail.

To @Davide Giraudo: A linearly independent subset is a subset that is not linearly dependent.

A subset of a vector space is called linearly dependent if there is an element in this subset which is a linear combination of a finite number of elements of this subset.

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    $\begingroup$ A Schauder basis need not be dense. It's the linear span of the Schauder basis that is dense. $\endgroup$ – Chris Eagle Aug 27 '11 at 11:05
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    $\begingroup$ While this is interesting information, I can't see how the non-existence of a Schauder basis in certain spaces has anything to do with the question. A Schauder basis is a much stronger concept than a linearly independent subset and given a linearly independent and dense subset I don't see how you should be able to extract or construct a Schauder basis. $\endgroup$ – t.b. Aug 27 '11 at 11:08
  • $\begingroup$ Thanks to the commenters. Perhaps, I need to get a better understanding of the subject. I will leave this answer for a while even it is not correct... $\endgroup$ – Yuri Vyatkin Aug 27 '11 at 11:29
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    $\begingroup$ It is a correct answer, but to the wrong question. $\endgroup$ – Mark Aug 27 '11 at 11:29
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    $\begingroup$ I should have given you a reference for Enflo's result. The original article is here, the only book I know which contains a detailed exposition (of a variant) is Lindenstrauss-Tzafriri, see section 2.d. A highly readable book on this aspect of the theory of Banach spaces is Albiac-Kalton. $\endgroup$ – t.b. Aug 28 '11 at 4:31

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