Why is a von Neumann algebra is closed with respect to weak * topology?

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I was trying to prove that the identity map between a von Neumann algebra $(A,\mbox{ultra weak topology})$ with respect to ultra weak topology and the von Neumann algebra $A$ with respect to weak* topology (A, weak* topology) is continuous $\mathrm{id} \colon (A,\mbox{ultra weak topology})\to (A, \mbox{weak*topology})$, where $A$ is von-Neumann.

Beside I know that these topologies are equivalent on $B(H)$ and since $A$ is weakly closed this implies that $A$ is ultra weakly closed Finally if i prove that $A$ is weak* closed then I can say that the identity map on $B(H)$ restricted on $A$, is continuous between $(A,\mbox{ultra weak topology})$ and $(A, \mbox{weak*topology})$.

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From your other question I know you are reading chapter 4 in Murphy's book. Because of the way Murphy defines the pre-dual $A_*$, the ultra-weak topology on $A$ is precisely the same topology as the weak$^*$. So there is nothing to prove.