# Neither open nor closed subspace of a vector space?

Consider a vector space $$V$$ over $$\mathbb{C}$$ with some norm (and topology induced by that norm). I am trying to find a subspace $$W \subset V$$ such that it is neither open nor closed in the topology mentioned above?

Motivation: I have studied that every banach space is essentially a closed subspace of $$\mathcal{C}(X)$$ for some compact Hausdorff space $$X$$ and hence I thought that we could also have some structure for normed linear spaces if the answer to my question is negative.

• When you say "subspace", do you mean in the linear algebraic sense of a vector space contained in $V$, or "subspace" in the topological sense of a subset equipped with the subspace topology? – Theo Bendit Feb 1 at 9:05
• @TheoBendit I meant it in a linear algebraic sense. – henceproved Feb 1 at 9:10
• You need $V$ to be infinite dimensional – Surb Feb 1 at 17:27

## 4 Answers

For a different flavor of example, consider $$C[0,1]$$ with the uniform norm. By a standard theorem (Stone-Weierstrass), the space $$P$$ of polynomial functions is dense in this space. But then, since $$P$$ is dense but clearly not equal to the whole space, it can't be closed in it.

Most of the standard Banach spaces you know are separable - which, in practice, means that they have a "nice" dense subspace. That dense subspace is a non-closed subspace.

You have to look to infinite dimensional spaces. In finite dimensional spaces, a subspace is always closed.

An example for an infinite dimensional space

There is a theorem stating that a linear form $$\varphi$$ is continuous if and only if its kernel $$W$$ is a closed subspace. And if $$\varphi$$ is discontinuous then the kernel is dense in $$V$$.

Hence if we find a non-continuous linear form, its kernel cannot be closed according to theorem above. And cannot be open either as an open subspace that is dense is the space $$V$$ itself.

Take $$V=\{P \in \mathcal C([0,1]), \mathbb C) \mid P \mbox{ is a polynomial}\}$$, equipped with the $$\sup$$ norm. The linear form $$\varphi: P \mapsto P(3)$$ is discontinuous. To see this consider the sequence $$P_n(x) = (x/2)^n$$. The sequence converges to zero for the $$\sup$$ norm but $$\lim\limits_{n \to \infty} \varphi(P_n) = \infty$$.

The kernel $$W$$ of $$\varphi$$ is the strange animal you're looking for!

Other answers have mentioned that it's impossible for a finite dimensional space, but I think it would be good to mention a proof:

If a space is finite dimensional, then it has a finite basis $$\{b_1, b_2, ... b_n\}$$. Given a sequence of vectors in the space $$v_k$$, each vector can be written as a linear combination of the basis vectors: $$v_k=\sum_{i=0}^n c_{i,k} b_i$$. Now suppose $$v_k$$ converges to some limit $$v$$. Then $$v = \sum_{i=0}^n c_i b_i$$. So $$\lim_{k \rightarrow \infty}\sum_{i=0}^n c_{i,k} b_i = \sum_{i=0}^n c_i b_i$$. Now $$\lim_{k \rightarrow \infty} c_{i,k}b_i$$ exists for each $$i$$, so we can then interchange the sum and the limit: $$\sum_{i=0}^n\lim_{k \rightarrow \infty} c_{i,k}b_i = \sum_{i=0}^n c_i b_i$$. If there is a subspace that contains $$v_k$$ for all $$k$$, then we can come up with a basis of the larger space such that the set of basis vectors of the subspace is a subset of the basis vectors of the larger space, and $$c_{i,k}$$ will be zero for any $$i$$ such that $$b_i$$ is not in the set of basis vectors of the subspace. Thus $$\lim_{k \rightarrow \infty} c_{i,k}$$ will be equal to zero for all such $$i$$, and $$v$$ will be a linear combination of only vectors that are in the set of basis vectors of the subspace, and thus will be a member of the subspace.

This shows that the limit of any sequence of vectors in the subspace will itself be in the subspace, and thus the subspace will be closed. So where did we use finite dimensionality? When we exchanged the sum and limit, that relied on the sum being finite. For infinite sums, the identity $$\sum_{i =0}^{\infty} \lim_{k \rightarrow \infty} a_{i,k}= \lim_{k \rightarrow \infty}\sum_{i=0}^{\infty} a_{i,k}$$ does not necessarily hold. To find a subspace that's not closed, we need an infinite dimensional space where this identity fails (or, as @mathcounterexamples.net mentions, find non-continuous linear forms, which is equivalent to the identity failing).

No proper subspace of a normed linear space can be open. Does that give you the answer? Additional facts: every infinite dimensional normed linear space contains a subspace which is not closed. In fact there exist discontinuous linear functional and their kernels are not closed.