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Consider the set $S=\{x\in\mathbb{R}:x^{2}<2\}$. Is it true that $\sup{(S)}$ must also be a real number? If so, what would be the proof strategy?? Is this also true for the infimum of $S$?

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    $\begingroup$ Please refer to the very definition of supremum (en.m.wikipedia.org/wiki/…). $\endgroup$ – Anurag A Dec 31 '14 at 17:32
  • $\begingroup$ The single most important property of the real numbers is that if a nonempty set of real numbers has an upper bound, then it has a supremum, which is also a real number. Since your set $S$ is nonempty and has an upper bound (every element is less than 1,000,000, for example) it must have a supremum. $\endgroup$ – MJD Dec 31 '14 at 22:03
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  • Is it true that $\sup S$ must also be a real number?

Let $S$ be a subset of $X$. By definition, the supremum is the least upper bound of the set $S$, and any upper bound lies in $X$. So $\sup S\in X$. In your case $X=\Bbb R$.

  • If so can you refer to a proof?

The statemente follows from the definition; see above answer.

  • Is this also true for the infimum of $S$?

The infimum is defined by $\inf S:=-\sup(-S)$, where $-S:=\{-x:x\in S\}$. So the answer is yes; see the first answer.


Definition 1 (Upper bound). Let $E$ be a subset of $X$, and let $M\in X$. We say that $M$ is an upper bound for $E$, iff we have $x\le M$ for every element $x$ in $E$.

The relation $\le$ is a order relation for the set $X$.

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$\Bbb R$ has the least upper bound property. Which says if $S\subset\Bbb R$ is a nonempty, set which is bounded above, then there is a real number $M$ such that $M=\sup S$.

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