# About sequentially complete and complete TVS

Let us use $X$ to mean a topological vector space. I know that if $X$ is a complete TVS then it is sequentially complete. I know that the converse is not true, so what I need now is to construct a TVS which is sequentially complete but not complete. I tried but cant find such space. Any help is very much appreciated...

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The dual of a separable and reflexive Banach space (so e.g. every $L^p$-space for $1<p<\infty$) equipped with the weak-star-topology is always not complete but sequentially complete. – Vobo Feb 19 '13 at 20:57

Let $X= \mathbb{R}^d$ with $d> \aleph_0$ and $H$ be the subspace of $X$ consisting of all vectors with countably many non zero coordinates.
First, $H$ is not complete. Indeed, in complete topological vector spaces any closed bounded ($A$ is bounded if for any neighbohood $V$ of $0$ there exists $r>0$ such that $A \subset rV$) subset is closed, but $H$ is dense in $X$ and $H \subsetneq X$ (since $d>\aleph_0$) so you can build a sequence $(x_n)$ in $H$ converging to some $x \in X \backslash H$: the closure of $\{x_n : n \geq 0\}$ in $H$ is a closed bounded non complete subset of $H$.
Then, you can show that $H$ is sequentially complete thanks to a diagonal argument.
Another mentionned example is $\ell^1$ with its weak topology.