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GLB=greatest lower bound LUB=least upper bound Give one example of a set such that the GLB and LUB exist but there exists at least one subset which has no GLB and LUB.

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    $\begingroup$ Hmm, is the set supposed to be a set of reals? The real-analysis tag suggests that it is, but boolean-algebra could imply that you're thinking about a more general order-theoretic setting. $\endgroup$ – Henning Makholm Jan 23 '17 at 12:44
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Consider the set $\{42\}$. It has a greatest lower bound and a least upper bound (both of which are $42$), but its subset $\varnothing$ has neither.

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  • $\begingroup$ A more non-trivial example? In the real line, this cannot happen non-trivially, right? $\endgroup$ – астон вілла олоф мэллбэрг Jan 23 '17 at 12:43
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    $\begingroup$ @астонвіллаолофмэллбэрг: Correct. If the original set has both upper and lower bounds, then all of its subsets are bounded, and so every non-empty subset will have a GLB and LUB. $\endgroup$ – Henning Makholm Jan 23 '17 at 12:45
  • $\begingroup$ Thank you for the reply, but what would be an example of an ordered set where this would be the case? That set certainly cannot have the GLB/LUB property, right? $\endgroup$ – астон вілла олоф мэллбэрг Jan 23 '17 at 12:47
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    $\begingroup$ @астонвіллаолофмэллбэрг: Obviously not -- but for example in $\mathbb Q$ we could consider $\{x\mid -2<x<2\}$ and its subset $\{x\mid x^2<2\}$. (That subset has a GLB and LUB in $\mathbb R$, of course, but not if $\mathbb Q$ is all we know). $\endgroup$ – Henning Makholm Jan 23 '17 at 12:51
  • $\begingroup$ Oh yes, that is correct. The GLB may exist, but we can not say it exists within the given set. That is a clear example. $\endgroup$ – астон вілла олоф мэллбэрг Jan 23 '17 at 12:55
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Consider the partially ordered set $$P=\{1,2,3,5,12,18,72,108,1080\}$$ ordered by divisibility. The set $X=\{5,12,18\}$ has greatest lower bound $1$ and least upper bound $1080,$ but its subset $Y=\{12,18\}$ has neither a greatest lower bound nor a least upper bound in $P.$

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