 # Let H be a Hilbert space and <mi mathvariant="script">A a commutative norm-closed unital Lorena Beard 2022-07-10 Answered
Let $H$ be a Hilbert space and $\mathcal{A}$ a commutative norm-closed unital $\ast$-subalgebra of $\mathcal{B}\left(H\right)$. Let $\mathcal{M}$ be the weak operator closure of $\mathcal{A}$.
Question: For given a projection $P\in \mathcal{M}$, is the following true?
$P=inf\left\{A\in \mathcal{A}:P\le A\le 1\right\}$
It seems that the infimum must exist and is a projection, but I am not able to show that the resulting projection cannot be strictly bigger than $P$. Also, if the above is true, what happens if $\mathcal{A}$ is non-commutative?
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Take $\mathcal{M}={L}^{\mathrm{\infty }}\left[0,1\right]$ and $\mathcal{A}=C\left[0,1\right]$.
Let $\mathbb{Q}\cap \left[0,1\right]={\left\{{r}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ be an enumeration of rationals in $\left[0,1\right]$, and define
$E:=\bigcup _{n=1}^{\mathrm{\infty }}\left({r}_{n}-\frac{ϵ}{{2}^{n}},{r}_{n}+\frac{ϵ}{{2}^{n}}\right)\cap \left[0,1\right]$
for some small $ϵ>0$ so that $m\left(E\right)$ is strictly less than $1$. Clearly, the only continuous function $f:\left[0,1\right]\to \left[0,1\right]$ satisfying ${\mathbb{1}}_{E}\le f\le 1$ is $f\equiv 1$, but the projection ${\mathbb{1}}_{E}$ is not equal to $1$.