> [!definition] > > Let $(\seqi{X}, \bracsn{\iota^i_j: i, j \in I, i \le j})$ be a [[Direct Limit|directed system]] of [[Topological Space|topological spaces]], then $(\bracs{X_i}, \bracsn{\iota^i_j})$ is **strict** if for each $i, j \in I$ with $i \le j$, the mapping $\iota^i_j: X_i \hookrightarrow X_j$ is a [[Topological Embedding|topological embedding]]. In which case, we identify $X_i$ as a subset of $X_j$, and $\iota^i_j: X_i \hookrightarrow X_j$ as the inclusion map. > > Let $(X, \bracsn{\iota^i_X})$ be the direct limit of this system, then > 1. $X$ can be identified with $\bigcup_{i \in I}X_i$. > 2. For each $i \in I$, $X_i$ can be identified as a subset of $X$, with $\iota_X^i: X_i \hookrightarrow X$ being the inclusion map. Moreover, $\iota^i_X(E) = X_i \cap E$ for all $E \subset X$, so the preimage operation can be identified with intersection with $X_i$. > 3. For each $i \in I$, $\iota_X^i: X_i \hookrightarrow X$ is a topological embedding. > > *Proof*. For any $x, y \in X_i$, since the inclusion maps are all injective, $x$ and $y$ are eventually equal if and only if $x = y$. Therefore $X$ can be identified with $\bigcup_{i \in I}X_i$ with $\iota_X^i: X_i \hookrightarrow X$ being inclusion maps. > > Let $i \in I$, and $U \subset X_i$ be [[Open Set|open]] in $X_i$. To show that $(\iota_X^i)^{-1}$ is continuous, it's sufficient to show that $\tau^i_X(U)$ is a relatively open in $X$. > > For each $j \in I$ with $j \ge i$, let $U_j \subset X_j$ be the largest open set such that $U_j \cap X_i = U$. Such an open set exists by taking the union of all open sets satisfying $U_j \cap X_i = U$. > > Let $k \in I$ with $i \le k \le j$, then $(\iota^k_j)^{-1}(U_j) = U_j \cap X_k$ is open in $X_k$, with $(\iota^k_j)^{-1}(U_j) \subset U_k$. Now, since $U_k \cap X_i = U$, there exists an open set $U_k^j \subset X_j$ such that $U_k^j \cap X_k = U_k$ and $U_k^j \cap X_i = U_k \cap X_i = U$, so $U_k \subset U_j$ as well. Therefore $U_j \cap X_k = U_k$ for all $k \in I$ with $i \le k \le j$. In particular, here the $U_j$s are ordered by their indices. > > Let $V = \bigcup_{j \ge i}U_j$, then $V \cap X_i = U$. Let $k \in I$, then there exists $k' \in I$ such that $k' \ge k , i$. Since the open sets are ordered, > $ > (\iota^k_X)(V) = X_k \cap V = X_k \cap \bigcup_{j \ge i}U_j = X_k \cap \bigcup_{j \ge k'}U_j = \bigcup_{j \ge k'}(\iota^k_{j})^{-1}(U_j) > $ > which is a union of open sets. Since $X_k \cap V$ is open for all $k \in I$, $V$ is an open set in $X$ such that $U = V \cap X_i$.