Character space of the unitization is one-point compactification

The algebraic correspondence. Put $K=\mathrm{\Delta }(A^{+})$ and $q={\chi }_{\mathrm{\infty }}$. For a character $\phi$ on $A$, define $E\phi (a,\lambda )=\phi (a)+\lambda$. By [F1], $E\phi ((a,\lambda )(b,\mu ))=\phi (a)\phi (b)+\lambda \phi (b)+\mu \phi (a)+\lambda \mu =E\phi (a,\lambda )E\phi (b,\mu )$. It is linear, unital, and nonzero. Conversely any $\chi \in K$ has $\chi (a,\lambda )=\chi (a,0)+\lambda$ by [F2]; its restriction to $A$ is either zero, giving $\chi =q$, or a character $\phi$, giving $\chi =E\phi$. Restriction is inverse to $E$ on $K\setminus \{q\}$, and $E\phi \mathrm{\ne }q$ because $\phi$ is nonzero. Contractivity of $E\phi$ and the norm extension give $\mathrm{\mid }\phi (a)\mathrm{\mid }\le \mathrm{\parallel }a\mathrm{\parallel }$, so the nonunital characters are bounded too.

Identify the topology on the complement first. For fixed $(a,\lambda )$, evaluation of $E\phi$ is the continuous function $\phi \mapsto \phi (a)+\lambda$. By the evaluation-topology definition [F2], $E:\mathrm{\Delta }(A)\to K$ is continuous. Its inverse on its image is restriction, whose evaluation at $a$ is the continuous function $\chi \mapsto \chi (a,0)$. Hence $E$ is a homeomorphism onto $K\setminus \{q\}$ with its subspace topology. Since $K$ is compact Hausdorff by [F2], it is locally compact (the whole space is a compact neighborhood of every point); its complement of the closed singleton $q$ is open and LCH by [F4]. Thus $\mathrm{\Delta }(A)$ is LCH without using any assumed compactification topology.

The point $q$ is not isolated. Under the isomorphism $\mathrm{\Gamma }:A^{+}\to C(K)$ of [F3], $\mathrm{\Gamma }(a,\lambda )(q)=\lambda$. Therefore $\mathrm{\Gamma }(A\oplus 0)$ is exactly the ideal $J=\{f\in C(K):f(q)=0\}$: one inclusion follows by evaluation, and for the reverse use surjectivity and the same equality. If $q$ were isolated, the function equal to zero at $q$ and one on its complement would be continuous and an identity for $J$. This ideal is nonzero because $A$ is nonzero, so its identity would be nonzero; its preimage would be a two-sided identity for $A$, contradicting genuine nonunitality. Hence $q$ is not isolated and $K\setminus \{q\}$ is dense. It is noncompact: otherwise its continuous image in Hausdorff $K$ would be closed by [F5], making $q$ isolated.

Compare all neighborhoods at infinity. Extend $E$ to a bijection $\mathrm{\Psi }:\mathrm{\Delta }(A{)}^{+}\to K$ by sending the added point to $q$. The two topologies already agree off infinity by step 2.1. If $U$ is open in $K$ and contains $q$, its complement $D=K\setminus U$ is compact by [F5] and contained in $K\setminus \{q\}$. The inverse homeomorphism in step 2.1 carries $D$ to a compact subset of $\mathrm{\Delta }(A)$, which is closed because that space is Hausdorff. Thus ${\mathrm{\Psi }}^{-1}U$ is open at infinity by [F4]. Conversely, if $C$ is closed compact in $\mathrm{\Delta }(A)$, then $E(C)$ is compact in Hausdorff $K$ and hence closed by [F5]; its complement is an open neighborhood of $q$ corresponding to $\mathrm{\Delta }(A{)}^{+}\setminus C$. This proves equality of the topologies and that $\mathrm{\Psi }$ is a homeomorphism. This argument also covers open sets containing both a character and infinity; no preimage is incorrectly confined to $\mathrm{\Delta }(A)$.

The zero case and conclusion. If $A=0$, [F1] gives $A^{+}=\mathbb{C}$. A nonzero complex-linear multiplicative functional on $\mathbb{C}$ is the identity: it has value one at 1 by [F2], so at $\lambda$ it has value $\lambda$. There are no nonzero linear functionals from the zero algebra. Hence $\mathrm{\Delta }(A)=\mathrm{\varnothing }$ and $K$ is the singleton, exactly ${\mathrm{\varnothing }}^{+}$. Its original subspace is not dense; density was asserted only in the nonzero genuinely nonunital case of step 2.2. In that case step 1.1 proves the displayed disjoint character decomposition, step 2.1 the complement homeomorphism and step 3.1 the one-point compactification with its weak-star topology.