C zero of a locally compact space

By [L0], $c_0(\mathbb{N})=C_0(\mathbb{N})$. This space is a nonzero commutative C*-algebra: completeness is [L1], pointwise multiplication and conjugation preserve null sequences, $\mathrm{\parallel }ab{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }b{\mathrm{\parallel }}_{\mathrm{\infty }}$, and $\mathrm{\parallel }{a}^{\ast }a{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }a{\mathrm{\parallel }}_{\mathrm{\infty }}^2$; it has no unit, since a unit $e$ would satisfy $e{\delta }_n={\delta }_n$ for every $n$, hence $e(n)=1$ for all $n$, contradicting $e\in c_0$. This proves claim 2 and licenses [L2].

Each ${\mathrm{e}\mathrm{v}}_n$ is a character of $c_0(\mathbb{N})$: it is complex-linear and multiplicative because evaluation at a point is, and it is nonzero because ${\mathrm{e}\mathrm{v}}_n({\delta }_n)=1$ for the coordinate vector ${\delta }_n\in c_0(\mathbb{N})$.

If $\chi$ is a character then $\chi ({\delta }_n)\in \{0,1\}$ for every $n$, because ${\delta }_n^2={\delta }_n$ gives $\chi ({\delta }_n{)}^2=\chi ({\delta }_n)$ and $\mathbb{C}$ is a field; the values are not all zero, since otherwise $\chi$ would vanish on all finite truncations by linearity and hence, by continuity from [L2] (licensed by [step 1.1]) and the density of truncations [L1], on all of $c_0$, contradicting that $\chi \mathrm{\ne }0$; and for $m\mathrm{\ne }n$ one has $0=\chi ({\delta }_m{\delta }_n)=\chi ({\delta }_m)\chi ({\delta }_n)$, so if $\chi ({\delta }_{n_0})=1$ then $\chi ({\delta }_m)=0$ for all $m\mathrm{\ne }{n}_0$.

For a character $\chi$ with $\chi ({\delta }_{n_0})=1$ and $a\in c_0(\mathbb{N})$ one has $\chi (a)={\sum }_n{a}_n\chi ({\delta }_n)=a_{n_0}$: approximate $a$ by its truncations [L1], use linearity on each truncation, and pass to the limit with continuity of $\chi$ from [L2]; the series has at most one nonzero term, because $\chi ({\delta }_n)=0$ for every $n\mathrm{\ne }{n}_0$ by [step 2.1], so the limit is $a_{n_0}\chi ({\delta }_{n_0})=a_{n_0}$ and no summability of $a$ is needed (a general element of $c_0$ need not be summable).

Hence every character is some evaluation, evaluations are characters by [step 1.2], and two evaluations are equal only if the indices agree, since ${\mathrm{e}\mathrm{v}}_m({\delta }_m)=1\mathrm{\ne }0={\mathrm{e}\mathrm{v}}_n({\delta }_m)$ for $m\mathrm{\ne }n$; this proves claim 1.

Claim 3: each ${\mathbf{1}}_F$ is a positive contraction, since ${\mathbf{1}}_F^2={\mathbf{1}}_F$ and $\mathrm{\parallel }{\mathbf{1}}_F{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$, and the family of finite subsets is directed by inclusion; for $a\in c_0$ and $ϵ>0$ choose $N$ with $\mathrm{\mid }{a}_n\mathrm{\mid }<ϵ$ for $n>N$ (possible because $a$ is a null sequence [L1]); then for every finite $F\supseteq \{0,\dots ,N\}$ one has $\mathrm{\parallel }{\mathbf{1}}_{F}a-a{\mathrm{\parallel }}_{\mathrm{\infty }}={\sup }_{n\mathrm{\notin }F}\mathrm{\mid }{a}_n\mathrm{\mid }\le {\sup }_{n>N}\mathrm{\mid }{a}_n\mathrm{\mid }<ϵ$.