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Let $V$ be a Banach space. Can you give me an example of a subspace $W\subset V$ (sub-vectorspace) that is not closed?

Can't find an example of that yet.

Thanks!

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  • $\begingroup$ Is every metric space Banach space? $\endgroup$
    – Paul
    Commented Aug 2, 2012 at 9:03
  • $\begingroup$ What Banach spaces do you know? Do you know a space of functions that is dense in another space of functions? Or do you know about $L^p$ or $\ell^p$-spaces? $\endgroup$
    – t.b.
    Commented Aug 2, 2012 at 9:06
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    $\begingroup$ @Paul-Banach spaces are normed and complete topological vector spaces. That's a way smaller class than metric spaces. $\endgroup$ Commented Aug 2, 2012 at 9:09
  • $\begingroup$ Sorry, I'm not very familar with this. $\endgroup$
    – Paul
    Commented Aug 2, 2012 at 9:11
  • $\begingroup$ @Kevin Carlson I see. Thanks for your comment. $\endgroup$
    – Paul
    Commented Aug 2, 2012 at 9:12

5 Answers 5

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There is a regular method to produce a lot of non-closed subspaces in arbitrary infinite dimensional Banach space.

Take any countable linearly independent family of vectors $\{w_i:i\in\mathbb{N}\}\subset V$ and define $W=\mathrm{span}\{w_i:i\in\mathbb{N}\}$. Then, $W$ is not closed.

Indeed, assume that $W$ is closed. Recall that $V$ is a Banach space, then $W$ is also Banach as closed subspace of Banach space. Linear dimension of $W$ is countable, but by corollary of Baire category theorem, Banach space can not have countable linear dimension. Contradiction, so $W$ is not closed.

This general result was demonstrated in Kevin's and J.J.'s answers. I'll show another one.

Consider Banach space $V=(C([0,1]),\Vert\cdot\Vert_\infty)$ of continuous functions with $\sup$ norm. Let $W=(P([0,1]),\Vert\cdot\Vert_\infty)$ be its proper subspace consisting of polynomials. It is of countable dimension because $W=\mathrm{span}\{x^k:k\in\mathbb{Z}_+\}$. From result given above it follows that $W$ is not closed.

But there is another proof. By Weierstrass theorem $W$ is dense in $V$, i.e. $\overline{W}=V\neq W$. Thus $W$ is not closed, thought it is dense in $V$.

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    $\begingroup$ how do you know that $W$ is a proper subspace of $V$, i.e., $V\neq W$? Can you show how by Weierstrass theorem $W$ is dense? $\endgroup$
    – Ruzayqat
    Commented Apr 23, 2016 at 20:38
  • $\begingroup$ I suppose this the result that was referred to: en.wikipedia.org/wiki/… $\endgroup$ Commented Dec 17, 2020 at 4:16
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A simple example, which gets at the difference between orthonormal and Hamel bases in infinite dimensions, is to take $H$ a separable infinite-dimensional Hilbert space and consider the span of its basis vectors $e_i, i \in \mathbb{N}$. This span, taken algebraically, certainly isn't the entire space because, for instance, $v=\sum_{i\in \mathbb{N}} \frac{1}{2^i} e_i$ isn't in it. Remember that spans are defined via finite linear combinations of the spanning set. But $v$ must be in the Hilbert space, because the partial sums define a Cauchy sequence and Hilbert spaces are complete.

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  • $\begingroup$ And how do you see that the span of the $e_i$ is not a closed subset? $\endgroup$
    – Sh4pe
    Commented Aug 2, 2012 at 9:14
  • $\begingroup$ @MattN. I see. I think too of course. Thanks MattN. $\endgroup$
    – Paul
    Commented Aug 2, 2012 at 9:25
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    $\begingroup$ @Sh4pe: I see from below you haven't applied the direct characterization of closedness: in any topological space a subset is closed iff it contains its limit points. In the special case of a subspace of a complete metric space, this is equivalent to completeness. $\endgroup$ Commented Aug 2, 2012 at 9:26
  • $\begingroup$ If that were true, wouldn't that imply that closure and completeness are the same? $\endgroup$
    – Sh4pe
    Commented Aug 2, 2012 at 9:28
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    $\begingroup$ You actually only need Hausdorff to get uniqueness for limits of sequences. Anyway the definition of limit point is more general than the limit of a sequence: $x$ is a limit point of $Y\subset X$ if every neighborhood of $x$ intersects $Y$. That shows why the above definition of "closed" works: if $Y$ is closed then its complement is a neighborhood of all its points, so $Y$ contains all its limit points. $\endgroup$ Commented Aug 2, 2012 at 9:32
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Take $V$ to be the space of sequences $(a_1,a_2,\dots)$ of real numbers with norm $\|(a_1,a_2,\dots)\| = \sum_{k=1}^\infty |a_1|$. Consider the subspace of $W \subset V$ consisting of sequences with only finitely many of $a_1,a_2,\dots$ being non-zero. Then $A_k = (1,\frac{1}{2},\dots,\frac{1}{2^k},0,0,\dots)$, $k=1,2,\dots$, gives a sequence of elements of $W$, but the limit of this sequence is not in $W$.

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  • $\begingroup$ This shows that $W$ is not complete. Is this equivalent to the fact that $W$ is not closed? The question came to my mind when I tried to understand the difference between completeness and closedness of subspaces. $\endgroup$
    – Sh4pe
    Commented Aug 2, 2012 at 9:16
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    $\begingroup$ @Sh4pe: A subspace is closed if and only if every converging sequence of points from the subspace has its limit in the subspace, so this shows that the subspace is not closed. More generally a complete subspace of a metric space is always closed, and a closed subspace of a complete space is complete. $\endgroup$
    – J. J.
    Commented Aug 2, 2012 at 9:40
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    $\begingroup$ @J.J. I can't understand well your example. You mean that the sequence $\{ A_k \}_k$ converges towards the identically zero sequence? If so, I can't understand how it is built $\endgroup$
    – VoB
    Commented Mar 9, 2019 at 18:23
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Take the space $L^1 [0,1]$. This space is complete.

Now take the subspace of continuous functions $C[0,1]$ (with the $L^1$ norm). Then this is not a Banach space (with respect to $\|\cdot\|_{L^1}$). To see this, consider the sequence

$$f_n(x) = \begin{cases} 0 & \text{on } \hspace{0.5cm} [0,\frac12 - \frac1n]\\ nx + 1 - \frac{n}{2} & \text{on } \hspace{0.5cm} [\frac12 - \frac1n, \frac12]\\ 1 & \text{on } \hspace{0.5cm} [\frac12,1]\\ \end{cases}$$

This is a Cauchy sequence but its limit is not continuous.

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    $\begingroup$ $f_n$ look more complicated than they are: $f_n$ is the function that is zero, then linear, then $1$. $\endgroup$ Commented Aug 2, 2012 at 9:41
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$C_{00}$ with sup norm is not closed in $l^{\infty}$

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