Proposition 9.5.5. Let $(\seq{E_n}, \seq{\iota_n^m|m, n \in \natp, m \le n})$ be a strict inductive system of locally convex spaces over $K \in \RC$ and $(E, \seq{\iota_n})$ be its direct limit, then:
For each $n \in \natp$, the topology of $E_{n}$ is induced by $\iota_{n}$, which allows the identification of $E_{n}$ as a subspace of $E$.
If $E_{n}$ is separated for all $n \in \natp$, then $E$ is also separated.
If $\iota^{m}_{n}(E_{m}) \subset E_{n}$ is closed for all $m, n \in \natp$ with $m \le n$, then the following are equivalent for any $B \subset E$:
$B$ is bounded.
There exists $n \in \natp$ such that $B \subset E_{n}$ is bounded.
If $E_{n}$ is complete for each $n \in \natp$, then $E$ is also complete.
Proof. (1): Let $U \in \cn_{E_n}(0)$. By Lemma 9.5.4, there exists $\bracs{U_m| m \in \natp, m \ge n}\subset 2^{E}$ such that $U_{n} = U$, $U_{m} \in \cn_{E_m}(0)$ and $U_{m}= U_{m + 1}\cap E_{m}$ for all $m \in \natp$. Let $V = \bigcup_{m \ge n}U_{m}$, then $V \cap E_{m} = U_{m}$ for all $m \ge n$. In particular, $V \cap E_{n} = U_{n} = U$.
(2): Let $x \in E \setminus \bracs{0}$, then there exists $n \in \natp$ such that $x \in E_{n}$. Since $E_{n}$ is separated, there exists $U \in \cn_{E_n}(0)$ with $x \not\in U$. By Lemma 9.5.4 and (1), there exists $V \in \cn_{E}(0)$ such that $V \cap E_{n} = U$, so $x \not\in V$.
(3), $\neg (b) \Rightarrow \neg (a)$: If $B \not\subset E_{n}$ for all $n \in \natp$, then there exists a subsequence $\bracsn{n_k}_{0}^{\infty} \subset \natp$ and $\seq{x_k}\subset B$ such that $x_{k} \in E_{n_{k}}\setminus E_{n_{k - 1}}$ for all $k \in \natp$.
Since $E_{n_k}\subset E_{n_{k+1}}$ is closed for all $k \in \natp$, there exists $\seq{U_k}\subset 2^{E}$ such that:
For each $k \in \natp$, $U_{k} \in \cn_{E_{n_k}}(0)$.
For each $k \in \natp$, $U_{k} = U_{k+1}\cap E_{n_k}$.
For each $k \in \natp$, $n^{-1}x_{k} \not\in U_{k}$.
then $V = \bigcup_{k \in \natp}U_{k} \in \cn_{E}(0)$ with $V \cap E_{n_k}= U_{k}$ for all $k \in \natp$. For any $n \in \natp$, $x_{k} \not\in nU_{k} = nV \cap E_{n_k}$. Therefore $B$ is not bounded.
(3), $(b) \Rightarrow (a)$: Let $U \in \cn_{E}(0)$, then $U \cap E_{n} \in \cn_{E_n}(0)$, so there exists $\lambda \in K$ with $\lambda (U \cap E_{n}) \supset B$.
(4): Let $\fF \subset 2^{E}$ be a Cauchy filter and
then $\fU$ is also a Cauchy filter, which converges if and only if $\fF$ does.
Since each $E_{n}$ is complete, it is sufficient to show that there exists $n \in \natp$ such that $F + U \cap E_{n} \ne \emptyset$ for all $F \in \fF$ and $U \in \cn_{E}(0)$.
Suppose for contradiction that for every $n \in \natp$, there exists $F_{n} \in \fF$ and $U_{n} \in \cn_{E}(0)$ such that $E_{n} \cap F_{n} + U_{n} = \emptyset$. Assume without loss of generality that for every $n \in \natp$, $U_{n}$ is convex and circled with $U_{n} \supset U_{n+1}$. Let
then since each $U_{n}$ is circled, so is $U$. Thus $U \cap E_{n} \supset U_{n} \cap E_{n} \in \cn_{E_n}(0)$, and $U \in \cn_{E}(0)$.
Now, suppose that $(F_{n} + U) \cap E_{n} \ne \emptyset$. Let $y \in (F_{n} + U) \cap E_{n}$, then there exists $N \in \natp$, $\bracs{x_k}_{1}^{N} \subset E$, $\bracs{\lambda_k}_{1}^{N} \subset [0, 1]$, and $z \in F_{n}$ such that
For each $1 \le k \le N$, $x_{k} \in U_{k} \cap E_{k}$.
$\sum_{k = 1}^{N} \lambda_{k} = 1$.
$y = z + \sum_{k = 1}^{N} \lambda_{k} x_{k}$.
In which case, since $U_{k}\supset U_{k+1}$ for all $k \in \natp$,
which is impossible. Therefore $(F_{n} + U) \cap E_{n} = \emptyset$ for all $n \in \natp$.
Finally, since $\fF$ is a Cauchy filter, there exists $F \in \fF$ such that $F - F \subset U$. Let $z \in F$, then there exists $n \in \natp$ such that $z \in E_{n}$. In which case, for any $y \in F \cap F_{n}$,
which contradicts the fact that $(F_{n} + U) \cap E_{n} = \emptyset$.$\square$