[proofplan] The [Schwartz topology](/page/Schwartz%20Space) is the locally convex topology generated by the family of semi-norms $\{\|\cdot\|_{\alpha,\beta}\}_{\alpha,\beta \in \mathbb{N}_0^n}$. By the [Neighbourhood Base of the Seminorm-Generated Locally Convex Topology](/theorems/664), a neighbourhood base at each point consists of finite intersections of semi-norm balls. The forward direction observes that each individual semi-norm ball is a neighbourhood, so topological convergence forces eventual containment. The converse takes an arbitrary neighbourhood, extracts the finitely many semi-norm conditions defining it, and uses pointwise semi-norm convergence to find a uniform index beyond which all conditions hold simultaneously. [/proofplan] [step:Establish the neighbourhood base structure of the Schwartz topology] The Schwartz semi-norms are the [functions](/page/Function) $\|\cdot\|_{\alpha,\beta}: \mathcal{S}(\mathbb{R}^n) \to [0,\infty)$ defined by \begin{align*} \|f\|_{\alpha,\beta} &:= \sup_{x \in \mathbb{R}^n} |x^\alpha \partial^\beta f(x)| \end{align*} for $\alpha, \beta \in \mathbb{N}_0^n$. The [Schwartz topology](/page/Schwartz%20Space) is the coarsest locally convex [topology](/page/Topology) making every $\|\cdot\|_{\alpha,\beta}$ [continuous](/page/Continuity). By the [Neighbourhood Base of the Seminorm-Generated Locally Convex Topology](/theorems/664), a neighbourhood base at each $f \in \mathcal{S}(\mathbb{R}^n)$ consists of all [sets](/page/Set) of the form \begin{align*} V_{(\alpha_1,\beta_1),\ldots,(\alpha_m,\beta_m);\,\varepsilon_1,\ldots,\varepsilon_m}(f) &:= \bigcap_{i=1}^{m} \{g \in \mathcal{S}(\mathbb{R}^n) : \|f - g\|_{\alpha_i, \beta_i} < \varepsilon_i\} \end{align*} for finitely many pairs $(\alpha_i, \beta_i) \in \mathbb{N}_0^n \times \mathbb{N}_0^n$ and positive reals $\varepsilon_1, \ldots, \varepsilon_m > 0$. [/step] [step:Show that topological convergence implies convergence in every semi-norm] Suppose $f_k \to f$ in the Schwartz topology. Fix $\alpha, \beta \in \mathbb{N}_0^n$ and $\varepsilon > 0$. The [set](/page/Set) \begin{align*} B_{\alpha,\beta,\varepsilon}(f) &:= \{g \in \mathcal{S}(\mathbb{R}^n) : \|f - g\|_{\alpha,\beta} < \varepsilon\} \end{align*} is a basic neighbourhood of $f$ in the Schwartz [topology](/page/Topology) (the special case $m = 1$ of the neighbourhood base from the [Neighbourhood Base of the Seminorm-Generated Locally Convex Topology](/theorems/664)). Since $f_k \to f$ in the topology, there exists $K \in \mathbb{N}$ such that $f_k \in B_{\alpha,\beta,\varepsilon}(f)$ for all $k \geq K$, which means $\|f_k - f\|_{\alpha,\beta} < \varepsilon$ for all $k \geq K$. Since $\varepsilon > 0$ was arbitrary, $\|f_k - f\|_{\alpha,\beta} \to 0$. [guided] We want to show that convergence in the Schwartz topology forces each individual semi-norm $\|f_k - f\|_{\alpha,\beta}$ to tend to zero. The key observation is that the single-semi-norm ball $B_{\alpha,\beta,\varepsilon}(f) = \{g : \|f - g\|_{\alpha,\beta} < \varepsilon\}$ is itself a basic neighbourhood of $f$ -- it is the special case $m = 1$ of the neighbourhood base provided by the [Neighbourhood Base of the Seminorm-Generated Locally Convex Topology](/theorems/664). Since $f_k \to f$ in the Schwartz topology, the [sequence](/page/Sequence) $(f_k)$ is eventually in every neighbourhood of $f$. In particular, for any fixed $\alpha, \beta \in \mathbb{N}_0^n$ and $\varepsilon > 0$, there exists $K \in \mathbb{N}$ such that $f_k \in B_{\alpha,\beta,\varepsilon}(f)$ for all $k \geq K$. Unwinding the definition of $B_{\alpha,\beta,\varepsilon}(f)$: \begin{align*} \|f_k - f\|_{\alpha,\beta} &< \varepsilon \quad \text{for all } k \geq K. \end{align*} Since $\varepsilon > 0$ was arbitrary, $\|f_k - f\|_{\alpha,\beta} \to 0$ as $k \to \infty$. [/guided] [/step] [step:Show that convergence in every semi-norm implies topological convergence] Suppose $\|f_k - f\|_{\alpha,\beta} \to 0$ for every $\alpha, \beta \in \mathbb{N}_0^n$. Let $U$ be an open neighbourhood of $f$ in the Schwartz [topology](/page/Topology). By the [Neighbourhood Base of the Seminorm-Generated Locally Convex Topology](/theorems/664), there exist finitely many pairs $(\alpha_1, \beta_1), \ldots, (\alpha_m, \beta_m) \in \mathbb{N}_0^n \times \mathbb{N}_0^n$ and positive reals $\varepsilon_1, \ldots, \varepsilon_m > 0$ such that \begin{align*} V &:= \bigcap_{i=1}^{m} \{g \in \mathcal{S}(\mathbb{R}^n) : \|f - g\|_{\alpha_i, \beta_i} < \varepsilon_i\} \subseteq U. \end{align*} By hypothesis, for each $i \in \{1, \ldots, m\}$, the convergence $\|f_k - f\|_{\alpha_i, \beta_i} \to 0$ provides $K_i \in \mathbb{N}$ such that $\|f_k - f\|_{\alpha_i, \beta_i} < \varepsilon_i$ for all $k \geq K_i$. Setting $K := \max(K_1, \ldots, K_m)$, every $k \geq K$ satisfies $\|f_k - f\|_{\alpha_i, \beta_i} < \varepsilon_i$ for all $i = 1, \ldots, m$ simultaneously, so $f_k \in V \subseteq U$. Since $U$ was an arbitrary open neighbourhood of $f$, the [sequence](/page/Sequence) $f_k$ converges to $f$ in the Schwartz topology. [guided] Now we prove the converse: if every semi-norm $\|f_k - f\|_{\alpha,\beta}$ tends to zero, then $f_k \to f$ in the Schwartz topology. The strategy is to take an arbitrary open neighbourhood $U$ of $f$ and show the sequence is eventually in $U$. By the [Neighbourhood Base of the Seminorm-Generated Locally Convex Topology](/theorems/664), $U$ contains a basic neighbourhood $V$ of $f$, which is a finite intersection of semi-norm balls: there exist finitely many pairs $(\alpha_1, \beta_1), \ldots, (\alpha_m, \beta_m)$ and tolerances $\varepsilon_1, \ldots, \varepsilon_m > 0$ with \begin{align*} V &:= \bigcap_{i=1}^{m} \{g \in \mathcal{S}(\mathbb{R}^n) : \|f - g\|_{\alpha_i, \beta_i} < \varepsilon_i\} \subseteq U. \end{align*} The crucial point is that this intersection involves only finitely many semi-norm conditions. For each $i \in \{1, \ldots, m\}$, the hypothesis $\|f_k - f\|_{\alpha_i, \beta_i} \to 0$ provides an index $K_i$ beyond which $\|f_k - f\|_{\alpha_i, \beta_i} < \varepsilon_i$. Taking the maximum $K := \max(K_1, \ldots, K_m)$ ensures all $m$ conditions hold simultaneously for $k \geq K$, so $f_k \in V \subseteq U$. Since $U$ was arbitrary, $f_k \to f$ in the Schwartz topology. [/guided] [/step]