# Metric and Topological structures induced by a norm

While proving that some normed spaces were complete, two questions came to my mind. They relate the topological and the metric structures induced by a norm.

1) Is it possible to find two equivalent norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on a vector space $V \$ such that $(V \ ,\|\cdot\|_1)$ is complete and $(V \ ,\|\cdot\|_2)$ is not?

2) Is there a vector space $V \$ and two non-equivalent norms such that $V \$ is complete relative to both?

Here I'm assuming $V \$ a vector space over a subfield of $\mathbb{C}$. Also I know that the answer is no if we only consider finite-dimensional vector spaces.

[Edit: I'm considering two norms equivalent if they define the same topology. I think it's the usual notion Jonas referred in his comment.]

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Regarding completeness, this notion actually only depends on $V$ as a topogolical vector space. (We can say that $v_n$ is a Cauchy sequence if for any neighbourhood $U$ of $0$, there is a natural number $N$ such that $v_n \in U$ if $n \geq N$.) This explains why the notion of Cauchy sequence, and hence the notion of completeness, is independent of any particular norm defining the given topology on $V$. –  Matt E Oct 12 '10 at 19:59
V has no chance of being complete unless it's a vector space over a complete subfield of C (so C or R), assuming the usual norm on C. –  Qiaochu Yuan Oct 12 '10 at 20:21
Dear Nuno, There is an intermediate notion between topological spaces and metric spaces, namely uniform spaces. These are spaces in which it makes sense to make uniform statements such as "for sufficiently large $m$ and $n$ the elements $x_m$ and $x_n$ are arbitrarily close", they key point being that we have a notion of what "close means" for any pair of points. (In a metric space we can do this, by asking that $d(x,y) < \epsilon$. In a general topological space, we can't do this: if we fix $x$ then we can use neighbourhoods of $x$ to say that $y$ is close to $x$, but we can't ... –  Matt E Oct 13 '10 at 0:02
... compare the sizes of neighbourhoods of different points, and so we can't have a notion of closeness which applies to a pair $x,y$ when both $x$ and $y$ are allowed to vary.) If $G$ is a topological group (e.g. a topological vector space) then it automatically becomes a uniform space, because we can measure closeness by choosing a neighbourhood $U$ of the identity, and then say that $x$ and $y$ are close if $x y^{-1}$ lies in $U$. (Here I am using multiplicative notation for the operation in $G$.) Another way to think of this is that we can use translation by group elements ... –  Matt E Oct 13 '10 at 0:05
... to move the neighbourhood $U$ around: e.g. $Uy$ is a neighbourhood of $y$, and $x$ and $y$ are "$U$-close" if $x \in Uy.$ In other words, in the presence of the group operation, which allows us to move points around, we can compare the size of neighbourhoods of different points. –  Matt E Oct 13 '10 at 0:06

1) No. It is straightforward to show that equivalent norms yield both the same convergent sequences and the same Cauchy sequences. (Written before Rasmus's answer was posted, but posted afterward.)

2) Yes. One way to see this is to note that isomorphism classes of vector spaces depend only on linear dimension, so the question amounts to finding 2 nonisomorphic Banach spaces of the same linear dimension. There are lots of examples of these. Every infinite dimensional separable Banach space has linear dimension $2^{\aleph_0}$. However, for example, $\ell^1$ and $c_0$ are separable Banach spaces that are not isomorphic (as Banach spaces).

Actually, "amounts to" wasn't quite accurate. It is certainly sufficient that the 2 Banach spaces are not isomorphic, but it is not necessary because you are only asking that one particular map (the identity in the original formulation) is not an isomorphism. So what I gave above is actually stronger. To just answer 2), you could just take any infinite dimensional Banach space and induce a new norm via an unbounded linear isomorphism with itself.

The answer above was assuming that equivalent norms are defined as in this PlanetMath article. If instead you meant only that the spaces are homeomorphic in the norm topologies, as Jyotirmoy Bhattacharya suspected, then the examples alluded to above won't work. However, there are also examples of pairs of Banach spaces that have the same linear dimension but are not homeomorphic, and this will work in either case. For example, $\ell^\infty$ and $c_0$ are not homeomorphic because $\ell^\infty$ is nonseparable. Both spaces have linear dimension $2^{\aleph_0}$. This was already mentioned for $c_0$, and for $\ell^\infty$ it follows because $c_0$ embeds in $l^\infty$ (which gives the lower bound on dimension) and because the cardinality of $\ell^\infty$ is $2^{\aleph_0}$ (which gives the upper bound).

(I'm now pretty sure this isn't what you want, based on your edit, but this still gives another example for the actual question as well as an answer to Jyotirmoy's comment.)

Incidentally, another way to see that $2^{\aleph_0}$ is a lower bound for the linear dimensions of $\ell^1$ and friends is to consider the linearly independent set $\{(1,t,t^2,t^3,\ldots):0\lt t\lt 1\}$. Cardinality of the spaces gives an upper bound.

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That was the easy part anyway. =) –  Rasmus Oct 12 '10 at 17:29
In (2), shouldn't we be looking at homeomorphisms since the OP is interested in topological equivalence? And isn't it then the case that all separable Banach spaces are homeomorphic to $l_2$ and hence to each other and therefore at least for separable spaces the answer to (2) is "No"? –  Jyotirmoy Bhattacharya Oct 12 '10 at 18:55
Good question. I assumed the usual notion of equivalent norms, and I don't think it is clear in the question. I'd ask for clarification, but I'll be away from computer for several hours starting now. –  Jonas Meyer Oct 12 '10 at 18:57
Thanks for this nice answer! About the first one, it was a lapse of memory. I forgot that property with constants characterizes equivalent norms. The other one I was expecting some counterexamples and I liked your approach. –  Nuno Oct 12 '10 at 19:35
@Jyotirmoy: Sorry I didn't fully understand your question, so would you mind reformulating it assuming that I'm not well-versed in functional analysis? –  Nuno Oct 12 '10 at 19:54