Can anyone give me an example of a vector space $V$ such that there is no norm which is complete on $V$?
Vector spaces with no Complete norms
-
0Do you assume $dim(V)<\infty$? – 2010-09-03
-
0(Every norm on a finite-dimensional vector space over a complete field is complete.) – 2010-09-03
-
5Hey, folks -- I'm not entirely sure this will fly, but I'll give it a go: I think that my answer contains enough information for Chandru1 to figure it out in a reasonably short amount of time, while still giving him something to think about. I think this is more educational for a beginning graduate student than just serving up the answer. So may I ask people to give Chandru1 at least a few hours to think things through by himself? – 2010-09-03
-
0@Pete: Yes, but I was not sure if also Chandru knows that. ;) – 2010-09-03
-
0@Pete L. Clark: Thanks a lot for you encouragement, and hints! – 2010-09-03
-
0@Chandru1: Please consider posting an answer once you figure it out. It could be helpful to others, and I think many of us will be positively reinforced by seeing the evidence of your work. – 2010-09-04
-
0@Pete L.Clark: Sure! I shall do it soon! – 2010-09-04
2 Answers
I assume you want $V$ to be a vector space over $\mathbb{R}$ or $\mathbb{C}$. Since the latter can be viewed as a special case of the former, let's say $V$ is an $\mathbb{R}$-vector space to fix ideas.
Hints:
1) A finite-dimensional subspace is always closed.
2) A proper closed subspace is nowhere dense.
3) There is a famous theorem about countable unions of nowhere dense subsets of a complete space.
Addendum: I decided that my comment calling for a lack of other answers after having already given an answer myself was something of a conflict of interest, so I have made this answer CW.
-
0I could prove (1). I am not able to show the second. (3) Are you talking about the Baire Category theorem. – 2010-09-03
-
0@Chandru1: (2) is equivalent to the fact that no proper subspace contains a nonempty open set. This comes down to the fact that if $B$ is any open neighborhood of the origin, then the span of $B$ (and even the set of all scalar multiples of $B$) is the whole space. (3). Yes, I am speaking of the Baire Category Theorem. – 2010-09-03
-
0You are pointing at that fact that an infinite dimensional banach space has dimension at least $\mathfrak{c}$. What can we say about spaces with dimension greater than $\mathfrak{c}$. Does there exist such a space with no complete norms? – 2016-09-30
-
1@Sahiba: Actually, all I said was that the dimension of a Banach space cannot be $\aleph_0$. Unless you assume the continuum hypothesis, you have to entertain the possibility of cardinals strictly in between $\aleph_0$ and $\mathfrak{c}$...but in fact no such thing can be the dimension of a Banach space: http://planetmath.org/banachspacesofinfinitedimensiondonothaveacountablehamelbasis. (Maybe you knew that already.) For cardinals greater than $\mathfrak{c}$: good question. I don't know. – 2016-09-30
Space $\mathcal{P}$ of all polynomials in one variable is not complete with respect to any norm as we know that dimension of an infinite-dimensional Banach space is at least $\mathfrak{c}$.
Edit: Proof of the fact that infinite dimensional vector space has dimension at least $\mathfrak{c}$
If not, then dim $X$ $< \mathfrak{c}$. So, $\exists$ a countable basis of $X$, say {$v_n$: n $\in \mathbb{N}$}. Let $X_n$ = span{$v_i$ : i=1,2,...,n}. Clearly, $X$= $\bigcup_{n \in \mathbb{N}}$ $X_n$. $X_n$ is finite dimensional subspace of $X$, hence closed. Claim: $\mathring{X_n}$ = $\emptyset$. If not, then $\exists$ x $\in$ $\mathring{X_n}$. Thus, $\exists$ r $>$ 0 such that $B(x,r)$ $\subseteq$ $X_n$. Let z $\in$ $X$ and y = $\frac{r}{2 \Vert z \Vert}$ z. Then $\Vert y-x \Vert$ = $\frac{r}{2} < r$. So, y $\in B(x,r) \subseteq$ $X_n$. Now, z=$\frac{2 \Vert z \Vert}{r}$ (y-x) $\in$ $X_n$ (since $X_n$ is a subspace). Hence, $X$ $\subseteq$ $X_n$. Thus, $X$ = $X_n$ (which is a contradiction since $X$ is infinite dimensional). Now, $\mathring{\overline{X_n}}$ = $\emptyset$. Therefore, $X_n$ is nowhere dense. Thus, $X$ is countable union of nowhere dense sets (which contradicts Baire Category Theorem). Hence, dim $X$ $\leq \mathfrak{c}$.
-
0I am not aware of this result, could you share a reference? – 2016-10-03
-
0It easily follows from Baire Category, however I have added the proof for your benefit. – 2016-10-03
-
0Thank you very much. By the way, there are norms on spaces of power series over a field, that make them Banach spaces, right? – 2016-10-03