Why isn't every linear subspace of an infinite-dimensional normed space closed?

Can somebody please point out the error in the following reasoning? I know that there exist non-closed subspaces of infinite-dimensional normed spaces.

Let $$V$$ be a vector space of infinite dimension equipped with a norm and $$U \subsetneq V$$ a proper subspace. $$U$$ has a basis $$\{\hat{u}_\alpha\}_{\alpha \in A}$$ and we can extend this basis with additional vectors $$\{\hat{w}_\beta\}_{\beta \in B}$$ so that $$\{\hat{u}_\alpha\} \cup \{\hat{w}_\beta\}$$ is a basis of $$V$$. Let's say that all these basis vectors are normed.

Now I want to prove that $$\partial U = U$$:
Let $$u$$ be a vector in $$U$$ and $$\beta \in B$$ arbitrary. For any $$\varepsilon > 0$$ the point $$u + \frac{\varepsilon}{2} \hat{w}_\beta$$ lies outside of $$U$$, so $$u \in \partial U$$.
Let $$v \in V \setminus U$$, so there exist a $$\beta \in B$$ and an $$a \in \mathbb{R}$$ so that $$a \,\hat{w}_\beta$$ appears in the linear combination that describes $$v$$. Now for every $$\varepsilon < |a|$$ the open ball of radius $$\varepsilon$$ around $$v$$ doesn't intersect with $$U$$. As a consequence $$v \not\in \partial U$$.
Since the boundary of a set is always closed, $$U$$ has to be closed.

• What may have lead you astray is that in infinite dimensional spaces we don't have any analogue of Gram-Schmidt orthogonalization method. Even assuming the space has an inner product (for example in a Hilbert space) Gram-Schmidt fails to produce uncountable sets of orthonormal vectors. That algorithm absolutely needs that sum over all the previous basis vectors in the recursive step. And without guaranteed convergence of infinite sums we're out of luck (or, the infinitely many correction terms may lead to a zero length vector). Therefore you cannot assume the extended bases to be orthogonal. – Jyrki Lahtonen Sep 23 '16 at 8:11
• Consider a Hilbert-space basis $\{e_n\}_{n\in N}$ for Hilbert space $H$ and the vector subspace $V$ of those $v\in H$ such that the inner product $v\cdot e_n=0$ for all but finitely many $n.$ Then $V$ is dense in $H$ but $V\ne H.$ – DanielWainfleet Sep 23 '16 at 12:08

Your mistake is here: "Now for every $\epsilon <|a|$..." Just because a vector has small norm (here $\epsilon$) its coefficients in a basis do not need to be small. In other words, the linear map sending a vector to one or more coefficients is not necessarily continuous.

You get some intuition for this by looking at two vectors in $R^2$ that are almost identical, yet independent. Then their difference has a small norm yet relatively large coefficients w.r.t. the basis of the two vectors. In infinite dimensions it can happen that successive basis vectors get more and more "almost dependent", thus giving rise to vectors that have larger and larger coefficients while having a norm bounded by a constant $\epsilon$.

To elaborate more on the relation of your question to continuity of linear maps, consider the map $L\colon V\to W$, $v=(u,w)\mapsto Lv:=w$, where $V=U\oplus W$ is a direct sum decomposition of $V$. If $L$ were continuous, then $U=\ker L$ would indeed be closed. Conversely, when $U$ is not closed, then $L$ is not continuous, which supplements the first paragraph.

• Your mistake was pseudo-answering this question. Could you elaborate, and better yet, don't stop at just concluding "wrong, sorry." I challenge you to write at least as much in the form of a hint, or suggesting a different approach, or by specifying how the OP can avoid the supposed mistake. – Namaste Sep 21 '16 at 18:05
• I upvoted this answer, because it does what the OP requested: "point out the error in the following reasoning." – Andreas Blass Sep 21 '16 at 18:18
• The OP "Wanted to know where the mistake was...is not equivalent to ":the OP wanted only to know where the mistake was". doesn't preclude expanding on an anwer. Note that I did NOT downvote your post; I was simply finding your answer mediocre, and thought you could easily improve it. – Namaste Sep 21 '16 at 18:37
• Usually when I answer questions on my phone I only give sketches and then answer follow ups in the comments. I like to tell myself this is beneficial to the learning experience. – Bananach Sep 21 '16 at 18:46
• I find this to be a perfectly fit answer to the question. – Mariano Suárez-Álvarez Sep 22 '16 at 3:12

Bananach's argument gets to the point. I want to add a bit about a method for identifying the error. Namely, I recommend that you test this kind of arguments against a case where you "know" that a subspace is not closed. Try it with your favorite! Below I walk you thru the first example that came to my mind. You will see that we end up exactly in a kind of situation Bananach described.

Let us consider the space $\ell^2$ of square summable real sequences (with the $L^2$-norm). It has as a subspace the set $V$ of sequences that eventually become constant zero. This subspace is spanned by the sequences $e_i, i=1,2,\ldots$, defined by declaring that the $i$th term of the sequence $e_i$ is equal to one, but the rest are all zero. You see that $||e_i||=1$. Let us then pick your favorite sequence from $\ell^2\setminus V$. For example we can use the sequence $x=(x_n)_{n>0}$ where $x_n=2^{-n/2}$. So $x_n^2=2^{-n}$ and therefore by the sum formula for a geometric series $||x||=1$ as well. We can just as well include $x$ in the (Hamel) basis of $\ell^2$ in addition to all the sequences $e_1,e_2,\ldots$.

You may already see where we are heading. Fix a (large) integer $k$. The linear combination $$x^{(k)}:=x-\frac1{\sqrt2}e_1-\frac1{\sqrt4}e_2-\cdots-\frac1{\sqrt{2^k}}e_k$$ looks like the sequence $x$ other than that the first $k$ entries are reset to zero. Therefore $||x^{(k)}||^2=2^{-k}$, so when $k\to\infty$ we see that the vectors $x^{(k)}$ become as short as we wish. We also see that $x-x^{(k)}$ is in the subspace $V$. So $x$ is in the closure $\overline{V}$.

But observe that in their respective vector space representations all the vectors $x^{(k)}$ share the coordinate $1$ of the basis element $x$ - even though they become arbitrarily short w.r.t. to our norm. Now reread Bananach's answer!

I might summarize this as:

• Infinite dimensional spaces are a bit tricky, and
• Hamel bases and coordinates w.r.t. them are not very useful for topological reasoning.

The latter is in sharp contrast to the finite dimensional case (where your intuition understandably comes from at this point of your studies).

There is a reason why people prefer to write elements of the example space $\ell^2$ as converging series using the "basis" elements $e_i$ as opposed to some Hamel basis (not that I could exhibit you a Hamel basis of $\ell^2$ in the first place - they only exist by virtue of Axiom of Choice).

• so basically the problem is that we can write a basis vector as the limit of a sequence of linear combinations of other basis vectors, right? – 0x539 Sep 23 '16 at 11:52
• Correct, @0x539 ! – Jyrki Lahtonen Sep 23 '16 at 12:41