I'm trying to show that if: $$ (a_{1n},a_{2n})\to (a_1,a_2)\\ (b_{1n},b_{2n})\to (b_1,b_2)\\ max\{a_{1n},a_{2n}\}\geq max\{b_{1n},b_{2n}\},\forall n\in\mathbb{N} $$ Then: $$ max\{a_1,a_2\}\geq max\{b_1,b_2\} $$
I tried proving it by showing that for $n$ big enough we'd have: $$ max\{a_1,a_2\}\geq max\{a_{1n}-\varepsilon,a_{2n}-\varepsilon\}\geq max\{b_{1n}+\varepsilon,b_{2n}+\varepsilon\}\geq max\{b_1,b_2\} $$ but I'm not sure if that is true, nor how to write it in a mathematically correct way.
Is the "continuity" conclusion true? Can anyone help me with this proof? Or is there a better way to show the result?
Thanks for helping! :DDDD
(1)Your second inequality is not valid.For example if $a_{1,n}=a_{2,n}=b_{1,n}=b_{2,n}=0$ for all $n$. ...(2)Proof by contradiction : If $$ B= \max(b_1,b_2)>\max (a_1,a_2) =A,$$ let $$r=B-A.$$ Then for all but finitely many $n$ we have : $$|a_{1,n}-a_1|<r/2,$$ $$ |a_{2,n}-a_2|<r/2,$$ $$ |b_{1,n}-b_1|<r/2,$$ $$ |b_{2,n}-b_2|<r/2,$$ but $B \in \{b_1,b_2 \}$ so one of these two must hold: $$ b_{1,n}>B-r/2$$ or $$b_{2,n}>B-r/2.$$ Hence $$\max (b_{1,n},b_{2,n}) >B-r/2$$ . Meanwhile $$a_{1,n}<a_1+r/2 < A+r/2 <B-r/2$$ and $$a_{2,n}<a_2+r/2 < A+r/2 < B-r/2$$. Therefore $$\max(a_{1,n},a_{2,n})<A+r/2<B-r/2< \max(b_{1_n},b_{2,n})$$ for all but finitely many $n$ , contradicting the conditions.