> [!definition] > > Let $E$ be a [[Vector Space|vector space]] over $K = \real|\complex$, equipped with a [[Topological Space|topology]] $\topo$. $E$ is a **topological vector space** if the addition map > $ > \cx \times \cx \to \cx \quad (x, y) \mapsto x + y > $ > and the scalar multiplication map > $ > K \times \cx \to \cx \quad (\lambda, x) \mapsto \lambda x > $ > are [[Continuity|continuous]]. [[Open Set|Open sets]] in $E$ are invariant under translation and multiplication. > [!definition] > > Let $E$ be a topological vector space. A [[Net|net]] $\net{x} \subset E$ is **Cauchy** if $\angles{x_\alpha - x_\beta}_{(\alpha, \beta) \in A^2}$ converges to $0$. > > $E$ is **complete** if every Cauchy [[Net|net]] in $E$ converges. > [!theorem] > > Let $E$ be a [[First Countable|first countable]] TVS, and every [[Cauchy Sequence|Cauchy sequence]] converges in $E$, then every Cauchy [[Net|net]] also converges in $E$. > > *Proof*. Let $\net{x}$ be a Cauchy net, then the net $\angles{x_\alpha - x_\beta}_{(\alpha, \beta) \in A^2}$ converges to $0$. > > ### Cauchy Sequence > > Let $\net{x}$ be a Cauchy net in $E$, then there exists $\seq{\alpha_k}$ such that $\seq{x_{\alpha_k}}$ is a Cauchy sequence. > > *Proof*. Let $\seq{U_n}$ be a countable [[Neighbourhood Base|neighbourhood base]] of $0$ such that $U_{n} \supset U_{n + 1}$ for all $n \in \nat$. > > Let $\alpha_1 \in A$. For any $n \in \nat$, choose $\alpha_{n + 1}$ such that $x_{\beta} - x_{\gamma} \in U_n$ for all $\beta, \gamma \gtrsim \alpha_{n + 1}$. Then $\seq{x_{\alpha_{n}}}$ is a Cauchy sequence, as $x_{\alpha_m} - x_{\alpha_n} \in U_{N}$ for all $m, n \ge N$. > > > ### Neighbourhood Reduction > > Let $\seq{U_n}$ be a countable neighbourhood base of $0$ such that $U_n \supset U_{n + 1}$ for all $n \in \nat$, then $\seq{2U_n}$ is also a neighbourhood base of $0$. > > *Proof*. Let $U \in \cn(0)^o$, then there exists $n \in \nat$ such that $U_n \subset \frac{1}{2}U$, and $2U_n \subset U$. > > > ### Control the Limit > > Let $x = \limv{n}x_{\alpha_n}$, then $x_\alpha \to x$. > > *Proof*. Let $\seq{U_n}$ be the same neighbourhood base as above, then $\seq{x + U_n}$ is a countable neighbourhood base for $x$ such that $x + U_n \supset x + U_{n + 1}$ for all $n \in \nat$, and $\seq{x + 2U_{n}}$ is also a countable neighbourhood base of $x$. > > Let $n \in \nat$, then there exists $N \in \nat$ such that $x_{\alpha_k} \in U_n$ for all $k \ge N$. Choose $N$ such that $N \ge n$, then $x_{\beta} - x_{\gamma} \in U_n$ for all $\beta, \gamma \gtrsim \alpha_{N}$, and > $ > x - x_\beta = \underbrace{x - x_{\alpha_N}}_{\in U_n} + \underbrace{x_{\alpha_N} - x_\beta}_{\in U_n} \in 2U_n \quad \forall \beta \gtrsim \alpha_N > $ > meaning that $x_\beta \in x + 2U_n$ for all $\beta \gtrsim \alpha_N$. Therefore $x_\alpha \to x$.