Completion of Linear Topological Spaces

Question

I would like to read the proof of the theorem stating that every linear topological space has a completion. Is it the same as arguments in metric spaces? Do you have a resource suggestion?

Answer 1

The proof given by Dietrich Burde in the comments is via a series of exercises, so if you want to do it that way, consider my answer as a spoiler.

You can find a proof in the classical book by F. Trèves: Topological vector spaces, distributions and kernels.

It's Theorem 5.2 on page 41.

Regarding your question on TVS' with a distance, see the exercise 8.5 on page 77.

It's not difficult to see that the author is trying to point us towards the function $f(x) = d(x_1, x_2)$ for $x = (x_1, x_2) \in E \times E$.

Answer 2

Consider the way we use Dedekind cuts on $\Bbb Q$ to get $\Bbb R.$ Apply this method (amost verbatim) on any linear order $L$ to get a linear order $L^*\supseteq L$ such that (i) $L$ is dense in $L^*$ in the order-topology of $L^*,$ and (ii) if $\emptyset\ne S\subset L^*$ then (iia) if $S$ has an upper bound in $L^*$ then $S$ has a lub in $L^*,$ and (iib) if $S$ has a lower bound in $L^*$ then $S$ has a glb in $L^*.$

My preferred version for the set $D$ of Dedekind cuts on $L$: By taking a suitable isomorphic copy of $L,$ we may assume WLOG that if $A,B\subset L$ then $(A,B)\not\in L.$ Let $L^*=L\cup D,$ where $D$ is the set of pairs $(A,B)$ with $A\cup B=L,$ with $A\ne \emptyset\ne B,$ with $\forall a\in A\,\forall b\in B\,(a<b),$ and such that $A$ has no largest member and $B$ has no least member.