Let $(\seq{E_n}, \bracs{T^m_n: m, n \in \nat, m \le n})$ be a [[Directed Set|directed system]] of [[Locally Convex Topological Vector Space|locally convex topological vector spaces]], and $(E, \seq{T^n_E})$ be its [[Direct Limit of Locally Convex Topological Vector Spaces|limit]]. Countable systems enjoy the ability of building open sets inductively, which allows preserving additional properties across limits. > [!theorem] > > If $(\seq{E_n}, \bracsn{T^m_n})$ is strict, then $T^n_E: E_n \to E$ is a [[Topological Embedding|topological embedding]] for each $n \in \nat$. > > *Proof*. Let $n \in \nat$ and $V \subset E_n$ be a [[Convexity|convex]], [[Balanced Set|balanced]], and [[Absorbing|absorbing]] [[Neighbourhood|neighbourhood]] of $0$. > > Let $V_n = V$. For each $m > n$, let $V_m \subset E_m$ be a convex, balanced, and absorbing neighbourhood of $0$ such that $(T^{m-1}_m)^{-1}(V_m) = V_{m - 1}$. From here, take $U = \bigcup_{m \ge n}V_m$, then for each $m \in \nat$, > $ > (T^{m}_E)^{-1}(U) = \bigcup_{k \ge m}(T^m_E)^{-1}(V_m) > $ > which is a convex, balanced, and absorbing neighbourhood of $0$, with $(T^n_E)^{-1}(U) = V_n = V$. > > [!theorem] > > Suppose that $(\seq{E_n}, \bracsn{T^m_n})$ is strict, each $E_n$ is [[Hausdorff Space|Hausdorff]], then $E$ is also Hausdorff. Moreover, if each $E_n$ is a closed subspace of $E_{n+1}$, then > 1. For each [[Sequence|sequence]] $\seq{x_n} \subset E$ and $x \in E$, $x_n \to 0$ if and only if there exists $N \in \nat$ such that $x_n \in E_N$ for all $n \in \nat$ and $x_n \to 0$ in $E_N$. > 2. For each [[Sequence|sequence]] $\seq{x_n} \subset E$ and $x \in E$, $x_n \to x$ if and only if there exists $N \in \nat$ such that $x_n \in E_N$ for all $n \in \nat$ and $x_n \to x$ in $E_N$. > > *Proof*. Let $x \in E$ with $x \ne 0$. Assume without loss of generality that $x \in E_1$ and let $U_1$ be a convex, balanced, and absorbing neighbourhood of $0$ in $E_1$ such that $x \not\in U_1$. > > From here, for each $n \in \nat$, let $U_{n + 1}$ be a convex, balanced, and absorbing neighbourhood of $0$ in $E_{n+1}$ such that $U_{n + 1} \cap E_n = U_n$. Take $U = \bigcup_{n \in \nat}U_n$, then $U$ is a convex, balanced, and absorbing neighbourhood of $0$ in $E$ that does not contain $x$. > > $(1)$: Suppose that no such $N \in \nat$ exists. By re-indexing, assume without loss of generality that $x_n \in E_n \setminus E_{n - 1}$ ($E_0 = \emptyset$) for all $n \in \nat$. > > Let $U_1$ be a convex, balanced, and absorbing neighbourhood of $0$ in $E_1$ that does not contain $x_1$. For each $n \in \nat$, let $U_{n + 1}$ be a convex, balanced, and absorbing neighbourhood of $0$ in $E_1$ such that $U_{n+1} \cap E_n = U_n$ and $x_{n+1} \not\in U_{n+1}$. Define $U = \bigcup_{n \in \nat}U_n$, then $U$ is a neighbourhood of $0$, but $x_n \not\in U$ for all $n \in \nat$.