$||T-T_n|| \rightarrow 0$ and $T_n$ are compact but $T$ is not a compact operator.

Question

It is a result that if $||T-T_n|| \rightarrow 0$ in the norm operator an that the $T_n \in \mathcal{L}(X,Y)$ (were $Y$ is a Banach space) are compact operators, then $T$ is compact. I found from here that pointwise convergence is not enough. But is there an easy counter-example for the case when $||T-T_n|| \rightarrow 0$ but $Y$ is not Banach and when then $T$ is not compact?

Answer 1

Consider $X=Y=d$ the space of finite sequences with the supremum norm. Then consider $$T_n(x_1,x_2,...,x_n,x_{n+1},...)=\left(\frac{x_1}{1},\frac{x_2}{2},...,\frac{x_n}{n},0,...\right).$$ This sequence of compact operators converges to $$T(x_1,x_2,...)=\left(\frac{x_1}{1},\frac{x_2}{2},...\right).$$ However this is not a compact operator since $T(B_1)$ is not complete.