2
$\begingroup$

Let $(E_\alpha,\tau_\alpha,g_\alpha)$ be a family of bornological (locally convex) topological vector spaces $(E_\alpha,\tau_\alpha)$, where a LCTVS $E$ is said to be bornological if every circled, bounded subset $A\subset E$ that absorbs every bounded set is a neighborhood of zero (i.e. has nonempty interior), and where $g_\alpha:E_\alpha\to E$ induces an inductive LCTVS structure on $(E,\tau)$.

Then in Shaefer's book Topological Vector Spaces, section II.8.2, we are told that $E$ must be bornological. However, I am slightly confused about a detail of the proof.

Let $A$ be a convex, circled subset of $E$ absorbing all bounded sets. If $B_\alpha$ is bounded in $(E_\alpha,\tau_\alpha)$ then $g_\alpha(B_\alpha)$ is bounded in $E$, then $A$ absorbs $g_\alpha(B_\alpha)$, whence $g_\alpha^{-1}(A)$ absorbs $B_\alpha$, and so $g_\alpha^{-1}(A)$ contains a neighborhood of $0$ in $E_\alpha$. Since this holds for all $\alpha$, then $A$ contains a neighborhood of $0$ in $E$.

However, I'm very confused about the last step — the inductive limit property does not directly give us the existence of opens in $E$, it merely places a constraint on them. So how do we figure out that $A$ contains a neighborhood of $0$ in $E$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

The topology on $E$ is such that it is the strongest (locally convex TVS) topology $\mathcal{T}$ such that for all $\alpha$: $g_\alpha: E_\alpha \to E$ is continuous. This implies that if $O \subseteq E$ is such that that $(g_\alpha)^{-1}[O]$ is open for all $\alpha$ then $O \in \mathcal{T}$. (this holds for convex circled sets, like $A$ per the comment by Jochen) For if this were not the case, we could add it to the topology of $E$ and generate a strictly larger LCTVS topology that still makes all $g_\alpha$ continuous. This explains the last sentence. Also see these notes for some more info, though Schaefer probably has more.

This is analogous to quotient maps (another example of a so-called final topology): $q: X \to Y$ is a quotient map (between topoloigcal spaces) iff $Y$ has the strongest topology that makes $q$ continuous iff $\forall O \subseteq Y: O \in \mathcal{T}_Y \Leftrightarrow q^{-1}[O] \in \mathcal{T}_X$, i.e. any set that doesn't contradict the continuity is in the topology already.

Improve this answer
$\endgroup$
5
  • 3
    $\begingroup$ One has to be careful: For the strongest locally convex topology making all $g_\alpha:E\alpha\to E$ continuous it is NOT TRUE that a subset $O$ is open if all pre-images $g_\alpha^{-1}(O)$ are open. This only holds for ABSOLUTELY CONVEX subsets $O$. $\endgroup$
    – Jochen Wengenroth
    Commented Jul 20, 2017 at 12:09
  • $\begingroup$ @JochenWengenroth interesting. What are absolutely convex sets? Convex only depends on linear structure ? $\endgroup$
    – Henno Brandsma
    Commented Jul 20, 2017 at 12:26
  • $\begingroup$ Absolutely convex means convex and circled. In many cases the locally inductive limit topology is in fact different from the inductive limit topology (in the category of topological spaces, category theorists usually speak of colimits), even for countable inductive limits with injective connecting maps (hence countable unions) of Banach spaces. See e.g. the book Barrelled locally convex spaces of Bonet and Perez Carreras. $\endgroup$
    – Jochen Wengenroth
    Commented Jul 20, 2017 at 12:41
  • $\begingroup$ @JochenWengenroth I find these inductive limits always hard to "grog", products I can understand ,or quotients too, but inductive limits seem to be a big thing in LCTVS theory. $\endgroup$
    – Henno Brandsma
    Commented Jul 20, 2017 at 12:55
  • 1
    $\begingroup$ They are not such a big thing. Perhaps you like the following: The topology of a LCS is given by a system of semi-norms. For the inductive limit you then just take all those semi-norms $p$ on $E$ such that all $p\circ g_\alpha$ are continuous on $E_\alpha$ (i.e., satisfy $p\circ g_\alpha \le c p_\alpha$ for some of the the given semi-norms on $E_\alpha$ and some $c$). Conceptually, this is very simple. However, in many cases it is not so easy to give a more "concrete" description of these semi-norms. $\endgroup$
    – Jochen Wengenroth
    Commented Jul 20, 2017 at 15:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .