Is there a $n \in \mathbb{N}$ such that for $f \in c_0$, we have $\alpha(f) = f(n)$? Prove or disprove.
Let $c_0$ be a subspace of $l^\infty$ consisting of sequences that converge to 0, and let $\alpha$ be a non-zero multiplicative functional on $c_0$. Show that there is a $n \in \mathbb{N}$ such that for $f \in c_0$, we have $\alpha(f) = f(n)$.
If anyone can offer any help, it would be very much appreciated. 
 A: By the Stone-Weierstrass theorem, if $X$ is a locally compact Hausdorff space (and $\mathbb{N}$ with the discrete topology is as such), then any closed subalgebra of $C_0(X)$  that separates points is either the entire space $C_0(X)$ or of the form $\{f\in C_0(X): f(x_0)=0\}$ for some $x_0\in X$.
Now let $\alpha$ be a non-zero multiplicative linear functional on $c_0=C_0(\mathbb{N})$. The kernel of $\alpha$ is a subalgebra of $c_0$. Moreover it is closed, since the functional is bounded. Also, the kernel of $\alpha$ separates points: Indeed, let $k,m\in\mathbb{N}$ be distinct integers. 
Suppose that we can't find a sequence $(x_n)\in\text{ker}(\alpha)$ such that $x_k\neq x_m$.
Note that $c_0/\text{ker}(\alpha)\cong \text{im}(\alpha)\leq\mathbb{R}$, therefore $c_0/\text{ker}(\alpha)$ is $1$-dimensional; Now the set $\{(x_n)+\ker(\alpha): x_k\neq x_m\}$ contains an infinite linearly independent subset (why? it is not hard to prove), which is a contradiction.
Conclusion: $\ker(\alpha)=c_0$ or there exists $n_0\in\mathbb{N}$ such that $\ker(\alpha)=\{(x_n)_n: x_{n_0}=0\}$. Since $\alpha$ is non-zero, the first case isn't possible.
