**Proof plan.** Each claim is verified by constructing a countable separating family of seminorms for the resulting space and checking completeness. The key observation in each case is that the relevant Cauchy condition for the new space can be translated into Cauchy conditions in the original space(s), where completeness is already known. **Step 1 (Closed subspaces).** [claim:Closed Subspaces Are Frechet] If $X$ is a Fréchet space with generating seminorms $\{p_n\}_{n=1}^\infty$ and $V \subseteq X$ is a closed subspace, then $V$ with the restricted seminorms $\{p_n|_V\}$ is a Fréchet space. [/claim] [proof] The restrictions $\{p_n|_V\}$ form a countable family of seminorms on $V$. They are separating: if $p_n(v) = 0$ for all $n$, then $v = 0$ in $X$, hence in $V$. The induced [topology](/page/Topology) on $V$ is the subspace topology (both are generated by the same seminorm balls intersected with $V$). For completeness: let $\{v_k\}$ be a [sequence](/page/Sequence) in $V$ that is Cauchy in every $p_n|_V$. Then $\{v_k\}$ is Cauchy in every $p_n$ on $X$, so by completeness of $X$, $v_k \to x$ for some $x \in X$. Since $V$ is closed and $v_k \in V$ for all $k$, the [limit](/page/Limit) $x$ belongs to $V$. [/proof] **Step 2 (Quotients by closed subspaces).** [claim:Quotients By Closed Subspaces Are Frechet] If $X$ is a Fréchet space with generating seminorms $\{p_n\}_{n=1}^\infty$ and $V \subseteq X$ is a closed subspace, then the quotient $X/V$ with the quotient seminorms $\hat{p}_n([x]) = \inf_{v \in V} p_n(x - v)$ is a Fréchet space. [/claim] [proof] Each $\hat{p}_n$ is a seminorm on $X/V$: absolute homogeneity follows from $\inf_{v \in V} p_n(\lambda x - v) = |\lambda| \inf_{v \in V} p_n(x - v/\lambda) = |\lambda| \hat{p}_n([x])$, and the triangle inequality follows from $\inf_{v} p_n((x+y) - v) \le \inf_{v_1, v_2} p_n((x - v_1) + (y - v_2)) \le \inf_{v_1} p_n(x - v_1) + \inf_{v_2} p_n(y - v_2)$. The family $\{\hat{p}_n\}$ is separating: if $\hat{p}_n([x]) = 0$ for all $n$, then for each $n$ there exist $v_n^{(k)} \in V$ with $p_n(x - v_n^{(k)}) \to 0$. In particular, taking a diagonal sequence, there exist $v_k \in V$ with $p_n(x - v_k) \to 0$ for every $n$ (this uses the countability of the seminorm family to extract a subsequence converging in all seminorms simultaneously). Hence $v_k \to x$ in $X$, and since $V$ is closed, $x \in V$, so $[x] = 0$. For completeness: let $\{[x_k]\}$ be Cauchy in every $\hat{p}_n$. We construct representatives converging in $X$. By the Cauchy condition, for each $j$ there exists $K_j$ such that $\hat{p}_n([x_k] - [x_{k'}]) < 2^{-j}$ for all $k, k' \ge K_j$ and all $n \le j$. Passing to a subsequence (still labelled $x_k$), we may assume $\hat{p}_n([x_{k+1}] - [x_k]) < 2^{-k}$ for all $n \le k$. By definition of the infimum, choose $v_k \in V$ with $p_n(x_{k+1} - x_k - v_k) < 2^{-k}$ for all $n \le k$. Redefine representatives: set $y_1 = x_1$ and $y_{k+1} = x_{k+1} - \sum_{j=1}^{k} v_j$. Then $[y_k] = [x_k]$ and $p_n(y_{k+1} - y_k) = p_n(x_{k+1} - x_k - v_k) < 2^{-k}$ for $n \le k$. For any fixed $n$, $\{y_k\}$ is Cauchy in $p_n$ (the tail $\sum_{k \ge N} p_n(y_{k+1} - y_k) < \sum_{k \ge N} 2^{-k} \to 0$ once $N \ge n$). By completeness of $X$, $y_k \to y$ for some $y \in X$, and $[x_k] = [y_k] \to [y]$ in $X/V$. [/proof] **Step 3 (Countable products).** [claim:Countable Products Are Frechet] If $\{X_j\}_{j=1}^\infty$ is a countable family of [Fréchet spaces](/page/Fr%C3%A9chet%20Space) with generating seminorms $\{p_n^{(j)}\}_{n=1}^\infty$ for each $X_j$, then $\prod_{j=1}^\infty X_j$ with the product topology is a Fréchet space. [/claim] [proof] Define seminorms on $\prod_j X_j$ by $q_{j,n}((x_1, x_2, \ldots)) = p_n^{(j)}(x_j)$ for each $j, n \in \mathbb{N}$. This is a countable separating family: $q_{j,n}(x) = 0$ for all $j, n$ implies $p_n^{(j)}(x_j) = 0$ for all $j, n$, so $x_j = 0$ for all $j$ (since each family $\{p_n^{(j)}\}$ is separating). The topology generated by $\{q_{j,n}\}$ is the product topology: convergence in every $q_{j,n}$ means coordinatewise convergence in every seminorm, which is precisely product convergence. Completeness: a sequence $\{x^{(k)}\}$ in $\prod_j X_j$ is Cauchy in every $q_{j,n}$ if and only if for each $j$, the sequence of $j$-th coordinates $\{x_j^{(k)}\}$ is Cauchy in every $p_n^{(j)}$. By completeness of each $X_j$, the $j$-th coordinate converges: $x_j^{(k)} \to y_j$ in $X_j$. Then $x^{(k)} \to y = (y_1, y_2, \ldots)$ in $\prod_j X_j$. [/proof] The three claims together establish all permanence properties. $\blacksquare$