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Let $X$ be a separable topological vector space with size (cardinal number) no larger than $\mathfrak{c}$. Does there exist any sequence of finite rank linear maps $\phi_n:X\to X$ pointwise converging to the identity mapping $id:X\to X$?

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  • $\begingroup$ Do you know whether this is true for, say, Banach spaces? $\endgroup$ – Jochen Glueck May 31 '18 at 14:21
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    $\begingroup$ I think that a separable Banach space without the approximation property would be a counterexample. I'll try to write down a proof in a couple of hours, but I wonder if there is something easier. $\endgroup$ – Nate Eldredge May 31 '18 at 14:27
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Just to chat, an easier counterexample is $X:=L^p(\mathbb{R})$ for $0\le p<1$, a complete metric separable TVS. The identity map can't be approximated by finite rank continuous linear operators, for the simple reason that there aren't any. The only convex open set of $X$ is $X$ itself. As a consequence, there aren't any non-zero continuous linear forms on $X$, nor any non-zero finite rank continuous linear operators.

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  • $\begingroup$ Great counterexample! $\endgroup$ – Nik Weaver May 31 '18 at 18:13
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I presume you want the operators $\phi_n$ to be continuous.

Let $X$ be a separable Banach space (which necessarily has cardinality $\mathfrak{c}$). If such a sequence exists, then $X$ has the approximation property. To see this, first note that we have $\phi_n \to id$ uniformly on compact subsets of $X$ (this is a nice exercise in the uniform boundedness principle - a good "prelim" problem). Now if $A : X \to X$ is a compact operator, it maps bounded sets inside compact sets, so $\phi_n A \to A$ uniformly on bounded sets, i.e. in operator norm. Thus $A$ is a norm limit of finite-rank operators.

However, not every separable Banach space has the approximation property. The first counterexample was constructed by Enflo in 1973. See the above-linked Wikipedia article for references.

It's quite possible that there is an easier counterexample.

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