Why is the image of the embedding $l^1 \to c_0$ not a closed subspace?

Question

My professor just noted this as a statement, that the image of the embedding $l^1 \to c_0$ is not a closed subspace. But he didn't tell why. Could someone explain why this is the case?

Answer 1

Without proof: $\ell^1\not\cong c_0$

Bounded inverse theorem: $$T:X\leftrightarrow Y:\quad\|T\|<\infty\iff\|T^{-1}\|<\infty$$

The range is dense: $$\iota:\ell^1\hookrightarrow c_0:\quad c_0=\overline{\ell_0}=\overline{\iota\ell_0}\subseteq\overline{\mathcal{R}\iota}\subseteq c_0$$

So if it were closed then
the above would imply that
their isomorphic as TVS.
Contradiction!