Some hint to prove that $\ell^{1}$ with $\ell^{\infty}$ norm is Banach space?

Question

I need show that $l^{1}$ with $l^{\infty}$ is normal space and Banach space.

For Normal space, I arguing about that $\ell^{1}$ norm and $\ell^{\infty}$ is equivalent.

For Banach space: I try show that $\ell^{1}$ is closed in $\ell^{\infty}$.

But am not sure the right way of proof. Please give me a hint about intuition how to prove them.

Answer 1

You can't do that, because $\ell^1$ is not closed in $\ell^\infty$. The sequence $x=\left(\frac1n\right)_{n\in\mathbb N}$ belongs to $\ell^\infty\setminus\ell^1$. But, if, for each $n\in\mathbb N$, $x_n$ is the sequence$$1,\frac12,\frac13,\ldots,\frac1n,0,0,\ldots,$$then $(\forall n\in\mathbb N):x_n\in\ell^1$ and $\lim_{n\to\infty}x_n=x$ in $\ell^\infty$.