I have tried to formulate the notion of products myself and this is what I came up with:

Let $(X_i, |*|_i), i\in I$ be a collection of normed linear spaces and $f_i:Y\to X_i$ a collection of bounded linear maps, all index by the set $I$.

As long as $I$ is a finit set we can factor all maps in the following way:

$Y\overset{\triangle}{\to}\prod_{I}Y\overset{\prod_{i\in I} f_i}{\to}\prod_{i\in I}X_i\overset{p_i}{\to}X_i$

where the norm on the products is given by:

$|x|=\underset{i\in I}{sup}(|x_i|_i)$

As can be seen this is a close relative to $l^\infty$. We can formulate the notion of coproducts in the dual way and get something similiar to $l^1$, but only with finit sums.

$|x|=\sum_{i\in I}|x_i|_i$

I originally choose the supremum norm (on the product) so that it would work with infinite products, I am, however, no longer sure.

I wanted to use the definiton above to illuminate the difference between the weak topology and weak convergence.

To start I imagine we are in the following position:

$X\overset{\triangle}{\to}\prod_{X'}X\overset{\prod_{\lambda\in X'} \lambda}{\to}\prod_{X'}\mathbb{R}\overset{p_i}{\to}\mathbb{R}$

The weak topology on $Y$ can be obtained as the coarsest topology where $(\prod_{\lambda\in X'}\lambda)\triangle$ is continuous with respect to the product topology in $\prod_{X'}\mathbb{R}$ while weak convergence correspondes to the product norm defined above, which correspondes to the box topology.

Now for the objects this seem to be working out fairly well but it occurs to me that $(\prod_{\lambda\in X'}\lambda)\triangle$ need not be bounded, infact as long as X' isn't uniformly bounded it won't.

So now I'm hoping that I did something wrong. I really want the category of normed vectorspaces to have infinit products and it makes me a bit sad to think it might not.

Alternatively this is why the ideas of uniform boundedness are so important and I need to incorporate these somehow. I read about them but I honestly didnt get thier significanse at the time and I can't really see thier place in the big picture. Any help in that regard would also be very appreaciated.

I apologice if the question is to vague to be a proper StackExchange question.


Your proposed construction does not describe a product in the category of normed spaces and bounded linear maps, and the issue is indeed a lack of uniform bounds.

There is a moral problem with the category of normed spaces and bounded linear maps, which is that isomorphism in this category does not capture isometric isomorphism of normed spaces. Any categorical constructions in this category are therefore only unique up to equivalence of norms.

The correct salvage is to change categories: you should be working in the category of normed spaces and weak contractions (bounded linear maps of norm at most $1$). Isomorphisms in this category are isometries, so categorical constructions in it are unique up to isometry. This category is both complete and cocomplete. Some details can be found in this blog post (which addresses the closely related category of Banach spaces and weak contractions). In particular, the categorical product of an arbitrary family $X_i, i \in I$ of normed spaces is the subspace of the set-theoretic product $\prod X_i$ for which the sup norm you describe exists, together with that norm.

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  • $\begingroup$ Nice blog post. I read the first chapter of hartshonrnes algebraic geometry once but I didn't really get the projective part so I dropped it. When I started reading FunkAna I couldnt help but notice similiarity to basic constructs in AlgGeo and this was pritty much the reason why I started thinking about categorical constructs with banach spaces. And now here I am again, facing the projective space in a different setting and I still dont have an intuitive idea of it.. You know of any other blogs for this particular ailment? $\endgroup$ – user25470 Jan 1 '13 at 14:52
  • $\begingroup$ @user: you might want to ask this as a separate question. $\endgroup$ – Qiaochu Yuan Jan 1 '13 at 21:04

Unfortunately (!?) the category of normed spaces does not have (even countable) infinite products. This is visible already in the product of countably many lines. This product exists as a pre-Frechet space, (meaning with a translation-invariant metric giving a locally convex topology) but definitely not as a normed space. (To prove that the topology on a metrizable space is not given by a norm, it suffices to show that some ball is not bounded... and interesting exercise here.)

Indeed, projective limits are closed subspaces of the corresponding products, so if countable products existed (which they don't), spaces such as $C^\infty[a,b]$ (the proj lim of $C^k[a,b]$) would have a topology given by a norm, which is not so. (One could say that it is unfortunate, but it is a fact.)

So, pre-Frechet spaces admit countable products.

Uncountable products, even of lines, will not be metrizable, because they will not admit a countable local basis. :)

Such issues/examples illustrate the need for a wider repertoire of topological vector spaces.

Edit: Indeed, as Qiaochu Y. notes, I did presume that the product of normed spaces would be a product of topological spaces. This raises an interesting point: does one want this, or not? If one can do without it, then Qiaochu's answer shows a suitably constrained category in which normed spaces do have products. But if one does need, or insists upon, the product being a product in the category of topological vector spaces, then there is none.

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  • $\begingroup$ Your argument seems to assume that the forgetful functor from normed spaces to topological spaces will preserve products. $\endgroup$ – Qiaochu Yuan Dec 31 '12 at 0:30
  • $\begingroup$ @Qiaochu Yuan: You can avoid this: Suppose $P$ is the product of countably many nonzero spaces $X_n$ with projections $p_n \colon P \to X_n$. The universal property of the product applied to the identity $X_n \to X_n$ and zero $X_n \to X_m$ shows that $p_n \neq 0$. Now apply the universal property of $P$ to $n p_n \colon P \to X_n$ to get a morphism $f \colon P \to P$ such that $n p_n = p_n f$ and this gives $n \lVert p_n \rVert = \lVert n p_n\rVert = \lVert p_n f\rVert \leq \lVert p_n\rVert \lVert f\rVert$, so $n \leq \lVert f\rVert$ for all $n$. No non-trivial infinite (co)products exist. $\endgroup$ – Martin Dec 31 '12 at 2:15

You might be interested in Lowen's approach spaces. Providing a pointer is about the limit of my competence for this subject.

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