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2 antoher typo

A particularly nice example that occurred naturally in my research is as follows:

Let $A$ algebra of $L^\infty$-equivalence classes of piecewise continuous functions on $[0,1]$ (or equivalently the algebra of PC functions with $f(x)$ equal to the right hand limit of $f$ at $x$ for each $x$) equipped with the uniform norm. Take our topological space $X$ to be the maximal ideal space/character space of $A$ (put the weak star topology on the set of multiplicative linear functionals).

We get the following nice properties:

$X$ is compact and Hausdorff (since it's the maximal ideal space of a commutative C*-algebra).

It's not metrizable because $A$ isn't separable; that metrizabilty of $X$ is equivalent to separability of $C(X)$ is a fun little exercise.

It's not too huge. The multiplicative linear functionals are just left limits, $x_l$, and right limits, $x_r$, at points $x$ in the interval so it has the same cardinality as $\mathbb R$.

$X$ is separable; let if $(x_n)$ are rationals tending to $x$ from the left then then ${x_n}_l$ tends to $x_l$ and likewise on the right.

A particularly nice example that occurred naturally in research is as follows:

Let $A$ algebra of $L^\infty$-equivalence classes of piecewise continuous functions on $[0,1]$ (or equivalently the algebra of PC functions with $f(x)$ equal to the right hand limit of $f$ at $x$ for each $x$) equipped with the uniform norm. Take our topological space $X$ to be the maximal ideal space/character space of $A$ (put the weak star topology on the set of multiplicative linear functionals).

We get the following nice properties:

$X$ is compact and Hausdorff (since it's the maximal ideal space of a commutative C*-algebra).

It's not metrizable because $A$ isn't separable; that metrizabilty of $X$ is equivalent to separability of $C(X)$ is a fun little exercise.

It's not too huge. The multiplicative linear functionals are just left limits, $x_l$, and right limits, $x_r$, at points $x$ in the interval so it has the same cardinality as $\mathbb R$.

$X$ is separable; let if $(x_n)$ are rationals tending to $x$ from the left then then ${x_n}_l$ tends to $x_l$ and likewise on the right.