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A von Neumann algebra $M$is a *-subalgebra of $B(H)$ which is closed in strong operator topology. In order to show that every such von Neumann algebra has a multiplicative unit, one employs the fact that $M$ (being a c-star-algebra) possesses an approximal unit. This sequence is positive, contained in the unit ball of $M$ and is increasing. Now, every source I know uses the fact that increasing nets of hermitian operators which is bounded above (that is $a_\lambda\leq b$ for some $b$ and for all $\lambda $converge strongly to a hermitian element (the supremum of the bespoken net).

Why does being contained in the unit ball of the algebra imply that it is bounded in the sense of the partial ordering ?

Especially one cannot use the identity operator of the underlying hilbert space, since showing that wlog we can assume that to be contained in M requires the existence of a multiplicative unit!

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Because $B(H)$ does. So your net is bounded in the unit ball of $B(H)$, and thus it is sot convergent there. But your von Neumann algebra is sot closed, so the limit stays in the algebra.

Another way of proving your assertion is to show that a von Neumann algebra is closed under arbitrary unions of projections (easy application of the Double Commutant Theorem). Then the union of all the projections in $M$ will be its unit. This requires that you first show that $M$ is spanned by its projections.

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