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Proposition: Given S is non-empty sup(S)=inf(S), prove that the set S has only one element.

What I stated, as this seemed rather trivial.

Proof: Let S be a non-empty subset of R that is bounded. It is given to us that sup(S)=inf(S). The claim is that S, then, has only one element within its set. We proceed by contradiction: Let a,b belong to S where a does not equal b and a is our smallest element, b is our largest element. It follows the inf(S) = a and sup(S) = b. It was given to to us sup(S)=inf(S) which entails that a = b showing uniqueness. Yet, we stated before that a doesn't equal b! We've reached a contradiction. Therefore, S contains only one element if sup(S)=inf(S).

Is this fine or does anyone think this could better stated?

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A minor thing: What if $S = \varnothing$? –  Johannes Kloos Feb 28 at 20:29
@JohannesKloos: It is often the convention to write $\sup\varnothing = -\infty$ and $\inf\varnothing = +\infty$, so this is still okay –  MPW Feb 28 at 20:41
@MPW: I know, but I wanted to point out a gap in the proof. –  Johannes Kloos Feb 28 at 20:48

2 Answers 2

up vote 3 down vote accepted

The proof is not right, but easily fixed. Let $a$ and $b$ be distinct elements of our set. Then $$\inf S\le a\lt b\le \sup S,$$ and we have our contradiction.

Remark: Note that $S$ need not have a smallest (or largest) element.

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If $\inf S = \sup S = b \in \bar{\mathbb R}$, you know that all elements of $S$ are at the same time $\geq b$ and $\leq b$. The statement follows trivially.

Note that it actually holds for all subsets of $\mathbb R$ if you define $\inf S = \infty$ and $\sup S = -\infty$ when $S$ is empty.

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I was actually going to write it this way earlier, but I was not sure how much depth was needed for the proof. I fixed the header at the top to only look at a non-empty set of R, as the original problem I got this from dealt with non-empty. Thank you for the insight though. –  Savage Henry Feb 28 at 20:44

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