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Let $\mathcal{X}$ be an abelian C* algebra without a unit. One can prove, that we can turn $\mathcal{X}_1 \equiv \mathcal{X} \bigoplus \mathbb{C}$ into an abelian, unital C*-algebra by defining the norm $\|\xi + \lambda\| \equiv \sup\{\|\xi x + \lambda x\| \colon x \in \mathcal{X}, \|x\| \leq 1\}$ for all $\xi+\lambda \in \mathcal{X}_1$ and the natural * -operation. Now we know, that since $\mathcal{X}_1$ is an abelian, unital C*-algebra, that the maximal ideal space $\mathcal{X}_1^* \supset \Sigma(\mathcal{X}_1)$ = all non-zero multiplicative, linear functionals $\mathcal{X}_1 \to \mathbb{C}$, endowed with the weak*-subspace topology, is a compact Hausdorff space. We furthermore note, that $\Sigma(\mathcal{X}_1) = \Sigma(\mathcal{X}) \cup \{\infty\}$, where $\infty(\xi + \lambda) \equiv \lambda$. This holds since for every $\infty \neq \phi \in \Sigma(\mathcal{X}_1)$ we have that $\phi$ is already determined by its values on $\mathcal{X} + 0 \subset \mathcal{X}_1$, and thus restricts to a unique element in $\Sigma(\mathcal{X})$. We then note, that $\Sigma(\mathcal{X}_1)-\{\infty\}$ is locally compact, since $\Sigma(\mathcal{X}_1)$ is compact, so in particular locally compact, and this is a property that is inherited by topological subspaces. Furthermore it is easy to prove, that the map $\Phi \colon \Sigma(\mathcal{X}) \to \Sigma(\mathcal{X}_1) - \{\infty\}$ given by $\phi \mapsto \hat{\phi}$, where $\hat{\phi}(\xi + \lambda) \equiv \phi(\xi) + \lambda$, is a homeomorphism, so $\Sigma(\mathcal{X})$ is also locally compact.

Now my question is if there is an easy way to see, that $\Sigma(\mathcal{X}_1)$ is the one-point compactification of $\Sigma(\mathcal{X})$, meaning that if $\mathcal{T}$ is the weak*-subspace topology on $\Sigma(\mathcal{X})$ then the topology on $\Sigma(\mathcal{X_1}) = \Sigma(\mathcal{X}) \cup \{\infty\}$ defined by $\mathcal{T_\infty} \equiv \mathcal{T} \cup \{(\Sigma(\mathcal{X}) - C) \cup \{\infty\} \mid C \subset \Sigma(\mathcal{X})$ is compact$\}$ coincides with the weak*-subspace topology on $\Sigma(\mathcal{X}_1)$?

h3fr43nd
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1 Answers1

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Note that $\Sigma(X)$ can be seen as a dense locally compact Hausdorff subspace of $\Sigma(X_1)$. Thus the result follows from this post: Showing one point compactification is unique up to homeomorphism

J. De Ro
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