System of inequalities with ordinal numbers

Question

Given a system of inequalities with ordinal numbers:

\begin{align} a_1 & >b_1 \\ & \,\,\,\vdots \\ a_n & >b_n \end{align}

Let $\phi$ be a permutation on $\{1,\ldots,n\}$. Is it true that $a_{\phi(1)}+\cdots+a_{\phi(n)}>b_1+\cdots+b_n$?

Answer 1

No. There are already counterexamples with $n=2$. Here is one: $\omega+1>\omega$ and $3>2$, but $3+(\omega+1)=(3+\omega)+1=\omega+1<\omega+2$.

Similar counterexamples can be obtained for any $n>1$: just consider $\omega+1>\omega$ and a bunch of inequalities of the form $k_i>m_i$ for $k_i,m_i$ finite, with the only requirement that $m_2+\dots+m_n>1$. Again, $k_2+\dots+k_n+(\omega+1)=\omega+1<\omega+m_2+\dots+m_n$.