Let $E$ be a Banach space, $Fr(E)$ be the set of all finite-rank operators from $E$ to $E$, $B(E)$ be the set of all continuous linear operators from $E$ to $E$. Then $Fr(E)$ is dense in $B(E)$ iff $dim(E)$ is finite.
I know that when $dim(E)$ is finite, $Fr(E) = B(E)$. The statement is true in this case. But I do not know how to prove the other side of this statement. There is a Hint: Use Riesz's theorem.If we start from the condition $Fr(E)$ is dense in $B(E)$, how can we derive $dim(E)$ is finite by using Riesz's theorem?
The deeper way to do this is to use Riesz Lemma to show that the unit ball is compact if and only if $\dim E<\infty$, and conclude that the identity operator is not compact, and thus not a limit of finite-rank operators.
But there is a much straightforward way. If $E$ is infinite-dimensional, no finite-rank operator can be invertible. This implies that $\|I-T\|\geq1$ for all finite rank $T$, so $I$ is not a limit of finite-rank operators.
The inequality is justified by the usual Neumann series trick, that if $\|I-T\|<1$, then $T$ is invertible. Namely, if $\|I-T\|<1$, then $\sum_{k=0}^\infty (I-T)^n$ is an inverse for $T$.