**Proof plan.** The strategy is to show that the subspace topology on $V_n$ from $V$ equals $\tau_n$, by proving the inclusion in each direction. The easy direction ($\tau_n$ is finer than the subspace topology) follows from the [continuity](/page/Continuity) of $j_n$. The hard direction (the subspace topology is finer than $\tau_n$) uses a layer-by-layer extension argument: given a $\tau_n$-neighbourhood $U_n$ of the origin, the strictness condition allows one to extend it inductively to a compatible system of neighbourhoods $U_n \subseteq U_{n+1} \subseteq \cdots$ whose union is open in $V$. **Step 1: The subspace topology is no finer than $\tau_n$.** By the definition of the [inductive limit topology](/page/Strict%20Inductive%20Limit%20Topology), the canonical injection $j_n : (V_n, \tau_n) \to (V, \tau_{\mathrm{ind}})$ is continuous. If $A \subseteq V$ is $\tau_{\mathrm{ind}}$-open, then $j_n^{-1}(A) = A \cap j_n(V_n)$ is $\tau_n$-open in $V_n$. This means every subspace-[open set](/page/Open%20Set) is $\tau_n$-open, so the subspace topology is coarser than (or equal to) $\tau_n$. **Step 2: Every $\tau_n$-neighbourhood extends to an inductive-[limit](/page/Limit)-open set.** [claim: Layer-By-Layer Extension] Let $U_n$ be a convex balanced neighbourhood of the origin in $(V_n, \tau_n)$. Then there exists a convex balanced set $U \subseteq V$ that is open in the inductive limit topology and satisfies $U \cap j_n(V_n) = j_n(U_n)$. [/claim] [proof] The proof constructs $U$ by inductive extension along the chain. For the pieces $V_1, \ldots, V_{n-1}$ below $V_n$, set $U_k := U_n \cap j_k(V_k)$ for $k < n$; since $j_k(V_k) \subseteq j_n(V_n)$ and $U_n$ is open in $(V_n, \tau_n)$, the [set](/page/Set) $U_k$ is open in the subspace topology of $j_k(V_k)$ in $(V_n, \tau_n)$, and the strictness condition (the topology $\tau_k$ coincides with the subspace topology from $V_{k+1}$, and hence from $V_n$ by induction) gives $U_k \in \tau_k$. For the pieces above $V_n$, proceed inductively. Suppose a convex balanced $\tau_k$-neighbourhood $U_k$ of the origin in $V_k$ has been constructed with $U_k \cap j_{k-1}(V_{k-1}) = j_{k-1}(U_{k-1})$. The strictness condition requires that $\tau_k$ coincides with the subspace topology from $(V_{k+1}, \tau_{k+1})$, so there exists a convex balanced $\tau_{k+1}$-neighbourhood $U_{k+1}$ of the origin in $V_{k+1}$ with $U_{k+1} \cap j_k(V_k) = j_k(U_k)$. This completes the inductive step. Define $U := \bigcup_{k=1}^\infty j_k(U_k) \subseteq V$. The set $U$ is convex and balanced because each $U_k$ is convex and balanced and the chain is compatible ($j_k(U_k) = U_{k+1} \cap j_k(V_k) \subseteq j_{k+1}(U_{k+1})$). For any $m \in \mathbb{N}$, \begin{align*} j_m^{-1}(U) &= j_m^{-1}\left(\bigcup_{k \geq m} j_k(U_k)\right) = U_m, \end{align*} where the second equality uses the compatibility $j_k(U_k) \cap j_m(V_m) = j_m(U_m)$ for $k \geq m$ (which follows from the inductive construction). Since $U_m$ is a $\tau_m$-neighbourhood of the origin for every $m$, the set $U$ is open in the inductive limit topology. By construction, $j_n^{-1}(U) = U_n$. [/proof] **Step 3: The subspace topology equals $\tau_n$.** Let $U_n$ be an arbitrary convex balanced $\tau_n$-neighbourhood of the origin. By Step 2, $U_n = j_n^{-1}(U)$ for some $\tau_{\mathrm{ind}}$-open set $U$. Hence $U_n$ is open in the subspace topology on $j_n(V_n)$. Since convex balanced neighbourhoods form a local base for any locally convex topology, every $\tau_n$-open set containing the origin is open in the subspace topology. Translating by an arbitrary point shows the two topologies agree everywhere. Combined with Step 1, $\tau_n$ equals the subspace topology, so $j_n$ is a [topological](/page/Topology) embedding.