Other than ∞, is there another case where a supremum (or an infimum for that matter) doesn't exist?
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$\begingroup$ Are you asking if a finite set always has a supremum? $\endgroup$– panciniJan 16, 2016 at 0:44
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1$\begingroup$ OP may rather be asking whether it is possible for a set bounded from above to lack a supremum. $\endgroup$– Brian TungJan 16, 2016 at 0:46
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3 Answers
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Supremum axiom: Any nonempty subset $A\subset\mathbb{R}$ which is bounded above has a supremum $s\in\mathbb{R}$.
This is only valid in $\mathbb{R}$. For example, the set $\{x\in\mathbb{Q}:x\geq 0, \ x^{2}<2\}$ does not admit supremum, since $\sqrt{2}\notin\mathbb{Q}$.
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Yes, depends on the partial ordering you define. A partial ordering for which a bounded set have sup and inf is called complete.
An example of an incomplete ordering would be on set $\lbrace a, b, c \rbrace$, $a < b$, $a < c$, but $b$ and $c$ are not comparable. Then the subset $\lbrace a, b, c \rbrace$ do not have a supremum
Note that the real numbers are complete, therefore supremum exists for all bounded sets
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1$\begingroup$ I think there might be something wrong with your example. You show that your partial order isn't a total order, but a complete order is different. You can have compete order without it being a total order for example $\{a,b,c,d\}$ with $a<b<d$ and $a<c<d$. $b$ and $c$ are not comparable, but every subset has a supremum(if both $b$ and $c$ are in the set, $d$ is the supremum). $\endgroup$ Jan 16, 2016 at 1:11
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Within the extended line $[-\infty,+\infty] = \mathbb R\cup \{\pm\infty\}$ every subset has a supremum and an infimum. Within the line $(-\infty,+\infty) = \mathbb R$ every subset has a supremum and and infimum except when the supremum or infimum within the extended line is $-\infty$ or $+\infty$. For example $$ \sup \{1,2,3,\ldots\} = +\infty $$ and $$ \sup\varnothing = -\infty. $$
