# find a $f \in (\ell^{1})^{*}$ so that $M \subseteq \ker f$

find a continuous functional $$f \in (\ell^{1})^{*}$$ so that $$M \subseteq \ker f$$

Note $$M:=\{x \in \ell^{c}: \sum\limits_{n \geq 1}x_{n}=0\}$$ and

$$\ell^{c}=\{x \in \ell^{\infty}:x_{n}=0 \operatorname{for all but finitely many} n \in \mathbb N\}$$

My only idea:

setting $$f(x)=\sum\limits_{n \geq 1}x_{n}$$ and we know for $$x \in M$$ that $$\sum\limits_{n \geq 1}x_{n}=0$$ thus $$M\subseteq \ker f$$

Is this correct?

Why does this suffice to show that $$M$$ is not dense in $$\ell^{1}$$

What you have done is correct. For the last part note the kernel of any continuous linear functional is closed. Hence, if $$M$$ is dense then it must be equal to $$\ell^{1}$$. This is certainly false, right?