# Why is an inductive limit of bornological spaces bornological?

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 $E$?

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.
• 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 $O$. – Jochen Wengenroth Jul 20 '17 at 12:09
• 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. – Jochen Wengenroth Jul 20 '17 at 15:01