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Motivation: I am doing functional analysis on locally convex spaces for the first time and I would like to know when I am allowed to characterise limit points and continuity sequentially. (This may well be a silly question.)

I phrase my question abstractly, but I mostly care about $\mathcal{C}_0^\infty(\mathbb{R}^n)$.

Let $(X_n)$ be an increasing sequence of first-countable locally convex topological vector spaces and let $X=\cup_n X_n$.

If we topologise $X$ with the finest locally convex topology such that the inclusions $X_n\rightarrow X$ are all continuous, is $X$ necessarily a first-countable space?

(A local base for such a topology is given by the the collection of all balanced, convex, absorbent sets whose intersections with every $X_j$ is open in $X_j$.)

I am currently reading from Reed & Simon Methods of Mathematical Physics, I cannot seem to find too many modern and systematic treatments of locally convex vector spaces!

goonfiend
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2 Answers2

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One can sharpen Alex Ravsky's answer:

Recall that a topological vector space is metrizable if and only if it is first countable.

Let $X_n \subsetneqq X_{n+1} \subsetneqq \cdots$ be a strictly increasing sequence of Fréchet spaces such that each $X_n$ carries the topology induced by $X_{n+1}$. Then $X = \varinjlim X_n$ is not metrizable.

The point is that $X$ is complete (see e.g. Schaefer-Wolf, Proposition 6.6 and its corollary). If $X$ were metrizable, then $X = \bigcup_{n=1}^\infty X_n$ would be a union of countably many nowhere dense sets, contradicting the Baire category theorem.

yup
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    Thankyou, I found and convinced myself of this result today. Fortunately we can often characterise continuity sequentially for such spaces in spite of the lack of countability, thanks to the following result:

    A linear map $T:X\rightarrow Y$ (where $X=\lim_{\rightarrow}X_n$ is given the inductive limit topology and $Y$ is locally convex), is continuous if and only if its restriction to every $X_n$ is continuous.

    Main point: If the spaces $X_n$ are first-countable (as they are in the case of the test functions), we can use sequences to characterise continuity of these restrictions.

     – goonfiend Apr 13 '13 at 09:48
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    @SeanGomes why are the $X_n$'s nowhere dense? –  Jul 28 '15 at 07:58
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It seems that usually $X$ is not first countable and even the direct limit $\mathbb{R}^\infty$ of $\mathbb{R^n}$ has the uncountable character, equal to the small cardinal $\mathfrak d$. If you need references, I shall ask my scientific consultant for them.

Alex Ravsky
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