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Suppose $(x_\alpha)_\alpha$ is an uncountable, linearly independent family of norm one vectors in a Banach space. Can one always select a basic sequence (or at least a minimal system) from this family? I suspect the answer is no but I cannot come up with an example.

Thank you!

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up vote 6 down vote accepted

Not a basic sequence. Consider $e_0 \oplus e_\gamma$ in $R\oplus H$ for $H$ a non separable Hilbert space, or, if you want a separable example, make $e_\gamma$ a Hamel basis for a separable Hilbert space.

EDIT: Aug 2. Every separated sequence of unit vectors contains a minimal subsequence with bounded biorthogonal functionals. There remains the case where your linearly independent set has compact closure. I have an example of such a set where any minimal sequence in the set has only unbounded biorthogonal functionals, but I do not know whether such a set must contain a minimal sequence.

Too bad you did not ask this a day earlier when we could have discussed it face to face. I'll write something down when I get a chance.

EDIT: Aug 2. It was not so bad to write--I was able to do it on the plane.

If $x_n$ is a separated subset of the unit sphere of $X$, then $x_n$ has a minimal subsequence whose biorthogonal functionals are uniformly bounded. Indeed, if $x_n$ does not have weakly compact closure, then it has a basic subsequence (see e.g. the book of Albiac-Kalton), so we can assume that $x_n\to x$ weakly. If $x=0$, then $x_n$ has a basic subsequence. If not, let $Q$ be the quotient map from $X$ onto $X/[x]$, where $[x]$ is the linear span of $x$. $Qx_n\to 0 $ weakly and is bounded away from zero by the separation assumption, hence has a basic subsequence $Qx_{n(k)}$, whence $x_{n(k)}$ is minimal with uniformly bounded biorthogonal functionals.

I don't know what can happen when $A$ is a linearly independent subset of the unit sphere of $X$ that is totally bounded, but there exists such sets so that every minimal sequence in the set has biorthogonal functionals that are not uniformly bounded. Consider the Cantor set as the branches of the infinite binary tree and let the nodes index the unit vector basis of $c_0$. Given a branch $t=(t(n))$ (where $t_1<t_2<\dots$ in the tree ordering) of the tree, let $x_t=\sum 2^{-n} e_{t(n)}$. By compactness (which is really just pigeonholing), any sequence $y_k$ of $x_t$-s has a subsequence that converges to some $x_s$, which means that for any $n$, if $k\ge k(n)$ then $y_k(j)=x_s(j)$ for $1\le j \le n$. From this it is easy to see that $y_k$ cannot be uniformly minimal.

In the above argument, try replacing the unit vector basis of $c_0$ with an appropriate normalized countably linearly independent sequence that has no minimal subsequence. I think it is known that such sequences exist. Probably an example is in Kadec's book. Maybe this will give an example that has no infinite minimal subset.

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Indeed, such sequences exist. Sequences that have the same closed span as any of their subsequences appear in the literature under the name overcomplete or overfilling (see for example Byorthogonal Systems in Banach Spaces, by Hajek, Montesinos, Vanderwerff and Zizler, Excercise 1.1, page 42). Every infinite dimensional separable Banach space contains such a sequence whose span is also dense. If $(x_n)_n$ is a set on the unit sphere whose span is dense in $X$, take, for example, $y_n=\sum_{k=0}^{\infty}\frac{x_k}{n^{k}k!}$. Then $(y_n)_n$ converges to $x_0$. Setting $z_n=n(y_n-x_0)$, we see that $(z_n)_n$ converges to $x_1$, and so on...Hence, any subsequence of $(y_n)_n$ has dense span.

Taking any normalized, $\omega$-linearly independent, overcomplete sequence and doing the previous construction on the binary tree suggested by Bill, indeed gives a linearly independent uncountable set that contains no infinite minimal subset.

Bill's answer was very helpful and completely answers my question. Thanks again.

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