1) Define −A = {x ∈ R; −x ∈ A}. If l is a lower bound for A, then −l is an upper bound for −A. Therefore, −A is upper bounded and, therefore, sup(−...

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1) Define −A = {x ∈ R; −x ∈ A}. If l is a lower bound for A, then −l is an upper bound for −A. Therefore, −A is upper bounded and, therefore, sup(−A) exists. We affirm that inf A = − sup(−A). In fact, since sup(−A) ≥ x, for all x ∈ (−A), then − sup(−A) ≤ −x, for all x ∈ (−A), ie, − sup(−A) ≤ y, for all y ∈ A. Therefore, − sup(−A) is a lower bound for A. If c > − sup(−A) is a lower bound for A, then −c < sup(−A) will be an upper bound for −A, which gives us a contradiction. Therefore, inf A exists and inf A = − sup(−A).
a) Define −A = {x ∈ R; −x ∈ A}.
b) If l is a lower bound for A, then −l is an upper bound for −A.
c) Therefore, −A is upper bounded and, therefore, sup(−A) exists.
d) We affirm that inf A = − sup(−A).
e) If c > − sup(−A) is a lower bound for A, then −c < sup(−A) will be an upper bound for −A, which gives us a contradiction.
f) Therefore, inf A exists and inf A = − sup(−A).

Ed · Answer

A alternativa correta é a letra d) "We affirm that inf A = − sup(−A)." (Nós afirmamos que inf A = - sup(-A)).

A explicação dada no texto mostra que, se l é um limite inferior para A, então -l é um limite superior para -A. Portanto, -A é limitado superiormente e, portanto, sup(-A) existe. Além disso, é afirmado que inf A = - sup(-A). Isso é verdade porque, se sup(-A) ≥ x para todo x ∈ (-A), então - sup(-A) ≤ -x para todo x ∈ (-A), ou seja, - sup(-A) é um limite inferior para A.

Assim, a alternativa d) é a correta.

1) Define −A = {x ∈ R; −x ∈ A}. If l is a lower bound for A, then −l is an upper bound for −A. Therefore, −A is upper bounded and, therefore, sup(−...

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