Proving inf A + inf B ≤ inf(A+B): Is it True?

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The discussion centers on the mathematical proof of the inequality inf A + inf B ≤ inf(A+B) for subsets A and B of real numbers. The user successfully demonstrated that inf A + inf B is less than or equal to inf(A+B) by utilizing the properties of infimum and the concept of ε (epsilon) to establish bounds. The proof involves showing that for every ε > 0, there exist elements a in A and b in B such that their sum is less than the sum of their infima plus a small margin. The conclusion affirms the validity of the inequality.

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  • Knowledge of ε (epsilon) arguments in mathematical proofs
  • Basic principles of inequalities in mathematics
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gauss mouse
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Hi, I'm just wondering this:
If A and B are subsets of the real numbers, is
inf A + inf B = inf(A+B)?

I've proved that
inf A + inf B <or= inf(A+B)?

I've tried the rest of the proof but it won't work. Is it even true?
 
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Let AI = inf(A), BI = inf(B) for every ε > 0, there exists a (in A) and b (in B) such that
a < AI + ε and b < BI + ε, a + b < AI + BI + 2ε, inf(A+B) < AI + BI + 2ε .

Since ε is arbitrarily small, inf(A+B) ≤ inf(A) + inf(B).
 
mathman said:
Let AI = inf(A), BI = inf(B) for every ε > 0, there exists a (in A) and b (in B) such that
a < AI + ε and b < BI + ε, a + b < AI + BI + 2ε, inf(A+B) < AI + BI + 2ε .

Since ε is arbitrarily small, inf(A+B) ≤ inf(A) + inf(B).
Thanks. I got it in the end but your way is much prettier.
 

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