**Proof plan.** The argument is by contradiction, using the [Baire category theorem](/theorems/630). If $V$ were metrizable, the [Completeness Theorem](/theorems/711) would make it a complete metrizable space and hence a Baire space. The decomposition $V = \bigcup_n V_n$ into closed subspaces with non-empty complement would then violate the Baire category theorem, since at least one $V_n$ would have non-empty interior, forcing $V_n = V$ and contradicting the strict inclusions. **Step 1: $V$ is a Baire space under the metrizability assumption.** Suppose for contradiction that $(V, \tau_{\mathrm{ind}})$ is metrizable. By the [Completeness Theorem](/theorems/711), $V$ is Hausdorff and complete. A complete metrizable space is a Baire space: the intersection of countably many dense [open sets](/page/Open%20Set) is dense, or equivalently, $V$ cannot be written as a countable union of [closed sets](/page/Closed%20Set) each with empty interior. **Step 2: Each $V_n$ is closed in $V$.** [claim: Closedness Of Each Piece In V] For every $n \in \mathbb{N}$, $j_n(V_n)$ is closed in $(V, \tau_{\mathrm{ind}})$. [/claim] [proof] The proof is by induction on the length of the chain above $V_n$. By the strictness assumption, $\iota_n(V_n)$ is closed in $(V_{n+1}, \tau_{n+1})$. By the [Embedding Property](/theorems/707), $j_{n+1}$ is a [topological](/page/Topology) embedding, so $j_{n+1}(\iota_n(V_n))$ is closed in the subspace topology of $j_{n+1}(V_{n+1})$ in $V$. Since $j_n(V_n) = j_{n+1}(\iota_n(V_n))$, this gives $j_n(V_n)$ closed in $j_{n+1}(V_{n+1})$. Repeating: $j_n(V_n)$ is closed in $j_m(V_m)$ for every $m > n$ (by induction, composing the closedness at each step). For any $v \in V \setminus j_n(V_n)$, the element $v$ lies in some $j_m(V_m)$ for $m > n$, and since $j_n(V_n)$ is closed in $j_m(V_m)$, there is a $\tau_{\mathrm{ind}}$-open neighbourhood of $v$ disjoint from $j_n(V_n)$. Hence $V \setminus j_n(V_n)$ is open and $j_n(V_n)$ is closed in $V$. [/proof] **Step 3: Baire category contradiction.** The space $V = \bigcup_{n=1}^\infty j_n(V_n)$ is a countable union of closed sets. Since $V$ is a Baire space (by the metrizability assumption and Step 1), at least one $j_{n_0}(V_{n_0})$ has non-empty interior in $V$: there exists a non-empty $\tau_{\mathrm{ind}}$-open set $U \subseteq j_{n_0}(V_{n_0})$. [claim: A Subspace With Interior Equals The Whole Space] If a linear subspace $E$ of a [topological vector space](/page/Topological%20Vector%20Space) $X$ contains a non-empty open set, then $E = X$. [/claim] [proof] Let $U \subseteq E$ be non-empty and open, and pick $e_0 \in U$. The [set](/page/Set) $U - e_0$ is an open neighbourhood of $0$ contained in $E$ (since $E$ is a subspace, $U - e_0 \subseteq E$). For any $x \in X$, the [continuity](/page/Continuity) of scalar multiplication gives $\lambda x \in U - e_0 \subseteq E$ for sufficiently small $\lambda > 0$. Since $E$ is a subspace, $x = \lambda^{-1}(\lambda x) \in E$. Hence $E = X$. [/proof] Applying this claim with $E = j_{n_0}(V_{n_0})$ and $X = V$ gives $j_{n_0}(V_{n_0}) = V$. But $V_{n_0} \subsetneq V_{n_0 + 1}$ implies $j_{n_0}(V_{n_0}) \subsetneq j_{n_0+1}(V_{n_0+1}) \subseteq V$, a contradiction. Therefore $V$ is not metrizable.