§4 — closed subspaces of quasi-complete spaces are quasi-complete

Let $\mathbb K$ be a real-or-complex scalar field (in practice $\mathbb R$ or $\mathbb C$), and let $V$ be a $\mathbb K$-vector space that is also a real vector space with compatible scalar multiplications (multiplying by a real number $r$ is the same as multiplying by $r\in\mathbb K$), carrying a topology in which addition and negation are continuous, scalar multiplication $\mathbb K\times V\to V$ is jointly continuous, and which is Hausdorff. Assume $V$ is quasi-complete over $\mathbb K$. Let $W$ be a $\mathbb K$-subspace of $V$ that is a closed subset of $V$. Then $W$, with the subspace topology and the induced operations, is quasi-complete over $\mathbb K$. Quasi-complete over $\mathbb K$ (the bundle's definition): for every index type $I$ (in the same universe as the space) carrying a partial order in which any two elements have a common upper bound, and every family $(v_i)_{i\in I}$ whose set of values is von Neumann bounded over $\mathbb K$ and which is Cauchy (for every neighbourhood $N$ of $0$ there is $i_0$ with $v_i-v_j\in N$ for all $i,j\ge i_0$), there is a point $x$ with $v_i-x\in N$ eventually (for all $i\ge$ some $i_0$) for every neighbourhood $N$ of $0$.