The question is whether there exits $\inf(A)$ such that $A$ is the union of all bounded subset of $R$ where each subset is closed under multiplication.
I think the infimum exists because completeness axiom states every non-empty bounded subset of $R$ has supremum and thus infimum. So I am trying to show that $A$ is also bounded and subset of $R$. Yet, I can not figure out how to show it's bounded.
Is my conclusion that $\inf(A)$ exists correct?