Absolute values of a closed set's elements

Let $B={|x|:xin A}.$ Suppose $bin bar B.$

There is a sequence $beta =(b_m)_{min Bbb N}$ of members of $B$ with $b=lim_{mto infty}b_m.$

And there is a sequence $alpha=(a_m)_{min Bbb N }$ of members of $A$ with $|a_m|=b_m$ for each $min Bbb N.$

Now $beta$ is a convergent series in $Bbb R$ so the set ${b_m:min Bbb N}$ is a bounded subset of $Bbb R.$ So there exists $Min Bbb R^+$ such that $|a_m|=b_mle M$ for each $min Bbb N.$

So ${a_m:min Bbb N}subseteq Acap {xin Bbb R^n: |x|le M}=^{def}C.$

Now $C$ is a closed bounded subset of $Bbb R^n$ (because $A$ is closed) so $C$ is compact. And $alpha$ is a sequence of members of $C.$ So $alpha$ has a subsequence $(a_{m_j})_{jin Bbb N}$ that converges to some $ain C.$ And $ain A$ (because $ain Csubseteq A)$.

We have $|;|a|-b_{m_j};|=|;|a|-|a_{m_j}|;|le |a-a_{m_j}|$ and we have $lim_{jto infty}|a-a_{m_j}|=0.$ Therefore $$b=lim_{jto infty}b_{m_j}=|a|in B.$$ So $bin bar Bimplies bin B.$ That is, $B$ is closed.

Remark: This will not work if you replace $Bbb R^n$ with an arbitrary complete normed linear space, where a closed bounded subset may fail to be compact. E.g. if $A={(1+2^{-n})e_n:nin Bbb N}$ where ${e_n:nin Bbb N}$ is an orthonormal Hilbert-space-basis for (infinite-dimensional) Hilbert space $H,$ then $A$ is closed in $H$, but ${|a|:ain A}={1+2^{-n}:nin Bbb N}$ is not closed in $Bbb R.$