Let $X$ be a Hausdorff space and let $C(X)$ denote the set of all continuous and bounded functions from $X$ to $\mathbb{C}$. It is a well-known fact that $C(X)$ forms a unital and abelian $C^*$-algebra (where the addition, multiplication, and scalar multiplication are defined pointwise, $C(X)$ is endowed with the sup norm, and the $*$ operation is conjugation)
Gelfand-Naimark theorem says the following: any unital and abelian $C^*$-algebra is isomorphic to $C(\mathbf{X}),$ where $\mathbf{X}$ is some compact and Hausdorff space, and $C(X)$ denotes the set of all continuous and bounded functions from $\mathbf{X}$ to $\mathbb{C}.$
Consequently, for any Hausdorff space $X,$ there exists a compact and Hausdorff space $\mathbf{X}$ such that $C(X)$ is isomorphic to $C(\mathbf{X}).$
Question: Is $\mathbf{X}$ a compactification of $X$?
By a compactification I mean the following: $\mathbf{X}$ is a compactification of $X$ if and only if $X$ embeds as an open and dense subspace in $\mathbf{X}$ (I know that the requirement that $X$ is open in $\mathbf{X}$ is usually not a part of the definition of a compactification, but I would like to include it)
Bonus question 1: What conditions do $C(X)$ or $X$ need to satisfy to make sure that $\mathbf{X}$ is a compactification of $X?$
Bonus question 2: If $C_b(X)$ denote the set of all bounded (and not necessarily continuous) functions from $X$ to $\mathbb{C},$ then $C_b(X)$ is a unital and abelian $C^*$ algebra. Is $\mathbf{X}$ from the Gelfand-Naimark theorem a compactification of $X?$ Why or why not?
As always, any help will be appreciated and rewarded.
Thank you.
Best Answer
In general (for any topological space $X$ at all), $\mathbf{X}$ is the Stone-Cech compactification of $X$. That is, there is a canonical continuous map $i:X\to\mathbf{X}$ such that for every compact Hausdorff space $K$ and every continuous map $f:X\to K$, there is a unique continuous map $g:\mathbf{X}\to K$ such that $gi=f$. (Explicitly, $\mathbf{X}$ is the space of $*$-homomorphisms $C(X)\to\mathbb{C}$, and $i:X\to\mathbf{X}$ takes any $x\in X$ to the homomorphism given by evaluation at $x$.)
Now, the Stone-Cech compactification is not always a compactification in your sense. Specifically, it is a compactification (embedding with dense open image) iff $X$ is locally compact and Hausdorff. This condition is obviously necessary, since any open subspace of a compact Hausdorff space is locally compact Hausdorff. Conversely, the image of $i$ is always dense, and $i$ is an embedding iff $X$ is completely regular. In particular, if $X$ is locally compact Hausdorff then $i$ is a embedding with dense image and its image is therefore open since any dense locally compact subspace of a compact Hausdorff space is open.
As for what you call $C_b(X)$, it is just $C(X_d)$ where $X_d$ is $X$ with the discrete topology. So, the $\mathbf{X}$ you obtain from this is just the Stone-Cech compactification of $X_d$. This will never be a compactification of $X$ in a canonical way unless $X$ is discrete so that $X=X_d$ (in particular, the canonical map $i:X\to\mathbf{X}$ is not continuous with respect to any topology on $X$ besides the discrete topology, since it is an embedding with respect to the discrete topology).
(Incidentally, what you call $C(X)$ is what is more commonly called $C_b(X)$. Typically $C(X)$ refers to the algebra of all (not necessarily bounded) continuous functions on $X$.)