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I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.

The Prompt:

Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x ∈ A$. Prove that $inf A = -sup( -A)$.

My Proof:

Define $B$ as the set of all lower bounds of $A$, and define $z∈B$ as the largest element of B, so $z=inf A$. For any $b∈B$, $a$ $\ge$ $b$ for all $a ∈ A$, so $0-b$ $\ge$ $0-a$ and $-b$ $\ge$ $-a$. Thus if we define the set $-B = \{-b|b∈B\}$, we can see that $-B$ is the set of upper bounds of $-A$. Since $z∈B$, $-z∈-B$, and since $b$ $\le$ $z$ for all $b∈B$, $0-b$ $\ge$ $0-z$ and $-b$ $\ge$ $-z$, so $-z$ is the smallest element of $-B$, and so $-z = sup(-A)$. Thus:

$inf A = z = - (-z) = -sup(-A)$

user10360304
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  • Note that you choose $z$ as the largest element of $B$. However is it allowed? If, yes, then you can just pick minimum element of $A$ rather than using sup and inf business. To sum up, the point of supremum and infimum is that for some sets there is no largest or smallest element, such as ${1/x: x \in \mathbb{N}}$. – ALNS Apr 19 '20 at 20:26

1 Answers1

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Mostly looks good. I would note though that you showed that $-B$ is a subset of the set of upper bounds of $-A$, and you really need it is the set of upper bounds, as claimed. This is easily dealt with though: suppose $x$ is an upper bound for $-A$. Then $x \geq -a$ for all $a \in A$, so $-x \leq a$ for all $a \in A$, so $-x$ is a lower bound for $A$, hence $-x \in B$, and $x \in -B$.

Physical Mathematics
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