# Locally convex inductive limit topology versus cofinal topology

Let $V$ be a complex vector space and let $\{0\} \subset V_1 \subset V_2 \subset \dots \subset V$ be an increasing sequences of subspaces of $V$, whose union is $V$. Suppose that each $V_n$ is a locally convex Hausdorff topological vector space and that every inclusion $V_{n} \subset V_{n+ 1}$ is continuous. We can then endow $V$ with two topologies:

1. The finest locally convex vector space topology, making all inclusions $V_n \subset V$ continuous (which can be shown to exist).
2. The finest topology on $V$ which makes makes all inclusion maps $V_n \subset V$ continuous (a general, point-set topology construction that works for all topological spaces).

Both of these topologies have universal properties characterizing continuous maps out of $V$: In the first case, this universal property holds within the category of locally convex spaces, while in the second, it holds in the category of all topological spaces.

My questions are:

1. Is there an example in which the second topology does even yield continuous vector space operations on $V$?
2. Does there exist an example in which the sceond topology does yield continuous vector spaces operations on $V$ but the two topologies are nonetheless different?

## 1 Answer

I think there would be problems with the "second" topology, as topological vector spaces do need to be Hausdorff and have continuous operations... I have not thought about that more general issue, but I have thought about the contrast between colimits and coproducts in the category of locally convex TVS's, and colimits and coproducts in the larger category of not-necessarily-locally-convex TVS's. (As expected, colimits are quotients of coproducts.)

Arbitrary locally convex TVS colimits exist, but uncountable not-necessarily-locally-convex TVS coproducts (and corresponding colimits) _do_not_exist_. E.g., an uncountable coproduct of copies of \C does not exist as a TVS. :)

One proof of the latter uses the non-local-convexity of the spaces $L^p$ with $0<p<1$ (defined slightly differently from the case of $p\ge 1$).