Is a subspace of a topological space always closed?

Question

Let $(X,T)$ be a topological space where $X$ has a vector space structure and let $V \subset X$.

Is it true that if $V$ is a vector subspace of $X$ then $V$ is closed in the topology $T$?

Sorry if this is very silly but I can't figure out a proof or counterexample. it is true in $\mathbb R^n$ for any $n \ge 1$ because lines, planes and hyperplanes passing in the origin are the subspaces and they are closed so I think we need an example from infinite dimensional spaces. I feel that the statement is not true

If it is not true for general topological spaces, is it true for particular types?

$R$ is a really bad example in topology. It's hugely atypical. — Stella Biderman, Apr 15 '16 at 13:48
It only makes sense to talk about vector subspaces of vector spaces. Do you mean "If $V$ is a (topological) subspace of $V$ with a vector space structure..."? — Travis Willse, Apr 15 '16 at 13:48
Also, closed where? Closed in the original space? It's always closed in the subspace topology — Stella Biderman, Apr 15 '16 at 13:49
It sounds like you want to start this question off something like this: "Let $X$ be a topological vector space..." https://en.wikipedia.org/wiki/Topological_vector_space — Lee Mosher, Apr 15 '16 at 13:53
Even in the case where $X$ is a vector space the subspace need not be closed; see, e.g., http://math.stackexchange.com/questions/177923/non-closed-subspace-of-a-banach-space . — Travis Willse, Apr 15 '16 at 13:54
For some subspaces that are very far from being closed (e.g. hyperplanes that are dense and not Borel), see Does there exist a linearly independent and dense subset?. — Dave L. Renfro, Apr 15 '16 at 14:02

Henno Brandsma · Accepted Answer · 2016-04-15T14:06:20.777

2

No, even for topological vector spaces (which is the only context this really makes sense).

Let $\mathbb{R}^\mathbb{N}$ be the topological vector space of sequences in the reals to themselves in the product (pointwise) topology.

Let $c_0$ be the subspace (linear as well) of all sequences that are $0$ except for at most finitely many terms.

Then $c_0$ is a linear subspace that is dense (and not closed) in $\mathbb{R}^\mathbb{N}$.

A finite-dimensional (!) linear subspace is always closed (assuming we are working over the reals or another complete field).

edited Apr 15 '16 at 14:06

answered Apr 15 '16 at 13:53

Henno Brandsma

242,131

The first thing I thought about for infinite dimensional case was $c_0$ but I couldn't verify that it is not closed. How to see that $c_0$ is not closed? – Apr 15 '16 at 13:55
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@madmatician The sequence $p_n$, where $p_n$ is a vector of $n$ ones followed by $0$'s lies all in $c_0$, while the pointwise limit is all $1$ and not in it. That $c_0$ is dense is clear from considering basic product open sets. – Henno Brandsma Apr 15 '16 at 13:57
Just one more: can you give me just a hint of how I can prove that a finite dimensional linear subspace is closed? – Apr 15 '16 at 14:00
E.g. here: http://mathoverflow.net/q/37849/2060 or http://math.stackexchange.com/a/1383132/4280 (norms are not really needed here). – Henno Brandsma Apr 15 '16 at 14:03

score 1 · Answer 2 · answered Apr 15 '16 at 14:17

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Here is a really accessible example : $X = \mathbb{R}$ is a $\mathbb{Q}$-vector space, and $ V = \mathbb{Q}\subset X $ is a sub-$\mathbb{Q}$-vector space which is not closed in $ X $.

answered Apr 15 '16 at 14:17

user293657

405

It also shows why we need a complete field for the finite-dimensional part to hold. – Henno Brandsma Apr 15 '16 at 16:34

Is a subspace of a topological space always closed?

2 Answers2