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Is there a way of defining representations of separable $C^*$-algebras, say $\Phi$, so that

  1. $\Phi(A)$ is faithful representation of $A$ on a separable Hilbert space.
  2. There is a closure operation $A\mapsto \overline{\Phi(A)(A)}^?$ so that when $A$ is commutative, $\overline{\Phi(A)(A)}^?$ is isomorphic to the space of bounded Borel functions on the spectrum of $A$?

EDIT: Thank you for your answer, however indeed, I had mind Borel functions mod null sets...

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  • $\begingroup$ Null sets generally mean sets of measure zero, which doesn’t mean anything without a measure being specified. You might want to consider instead meager sets, which are countable unions of nowhere dense sets, which is defined using only the topology. $\endgroup$
    – David Gao
    Commented Oct 16, 2023 at 15:18

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Assuming the closure operation you want still gives you a subalgebra of the bounded operators on the same separable Hilbert space, this is impossible in general, since, for example, the algebra of bounded Borel functions on $[0, 1]$ (which is, of course, the spectrum of $C([0, 1])$) doesn’t admit any faithful representation on a separable Hilbert space, as this algebra contains an uncountable collection of mutually orthogonal nonzero projections (say, the projection onto each point in $[0, 1]$).

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As requested, my comments below this answer are incorporated here: If, instead of the algebra of all bounded Borel functions on the spectrum of $A$, we instead want the algebra of bounded Borel functions modulo functions supported on a meager set, and by “closure operation” we mean a closure operator, then as the said algebra is the injective envelope of $A$, we can take the closure of $\Phi(A)$ to be the intersection of all injective subalgebras of $\mathbb{B}(H)$ containing $\Phi(A)$.

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    $\begingroup$ Though if you want the algebra of bounded Borel functions on the spectrum modulo functions supported on a meager set, then that is the injective envelope of $A$, so you can take the minimal injective subalgebra containing $\Phi(A)$. $\endgroup$
    – David Gao
    Commented Oct 15, 2023 at 23:18
  • $\begingroup$ @user52345435 In that case, if by “closure operation” you mean a closure operator (there’s a definition on Wikipedia if you haven’t seen it before), then you can take the intersection of all injective subalgebras of $\mathbb{B}(H)$ containing $\Phi(A)$. $\endgroup$
    – David Gao
    Commented Oct 16, 2023 at 15:16

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