Proving σmax(A-B) ≤ σmax(A) - σmin(B)

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SUMMARY

The inequality σmax(A-B) ≤ σmax(A) - σmin(B) is established as true based on numerical verification. The proof involves recognizing that for any element a in set A, it holds that a ≤ σmax(A), and for any element b in set B, it follows that b ≥ σmin(B). Consequently, the expression a - b is bounded by the maximum of A minus the minimum of B, confirming the inequality holds for all elements in the respective sets.

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hayu601
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Is it true if I state:

σmax(A-B) <= σmax(A) - σmin(B) ?

I verify numerically that it is correct but how to prove it?
 
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If a\in A then a\le max(A)
If b\in B then b\ge min(a) so -b\ge -min(B)

Adding, a-b\le max(A)- min(B) for all a in A, b in B/
 

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