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I need to show minimum, maximum, infimum and supremum, if they exist.

$$ C:= \bigcup_{n \in \mathbb{N}} [0,1/n[.$$

The Archimedean property says: let $e$, $x$ be real numbers, if $e>0$ and $x>0$ then there exists $n\in \mathbb{N}$ such that $ne>x$.

I cannot start anything with what i know form these statements, how can I show whether the statement above has a min, or max, inf or sup?

Xander Henderson
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doniyor
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3 Answers3

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Hint: $$ C:= \bigcup_{n \in \mathbb{N}}\; [0,1/n[ \;\;=\;\; [0, 1[ \;\;= [0, 1)$$

This is because $C = \left[0, \frac11\right) = \left[0, 1\right)$ contains every half-open interval of the form: $\left[0, \frac1n\right),\;\;n\geq 1$.

Put differently $$\left[1, \frac1n \right)\subseteq \left[0, \frac11\right) =\left[0, 1\right) = C \;\;\forall n \in \mathbb{N}.$$

You can use the Archimedean Property to establish whether or not $C$ has a maximum. The supremum exists; you need to determine that value (again, archimedean property to the rescue), but you also need to determine whether a maximum exists. If $C$ contains a maximum value, then that maximum must equal the supremum. Otherwise, there is no maximal element.

amWhy
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  • yeah, thanks. but how do i know which value is correct. there are endless numbers which live between 1 and 1/n ? – doniyor Dec 01 '12 at 14:32
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HINT: $$[0,1)\supseteq\left[0,\frac12\right)\supseteq\left[0,\frac13\right)\supseteq\left[0,\frac14\right)\supseteq\ldots$$

Brian M. Scott
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  • so the min and inf is $0$ and max and sup is something that converges against $0$ ? – doniyor Nov 30 '12 at 22:40
  • @doniyor: You’re right that $\inf C=\min C=0$, but the rest of what you said there makes no sense: $\frac34\in C$, so clearly both $\max C$, if it exists, and $\sup C$ must be at least $\frac34$. $C$ is actually an interval of some kind; what interval is it? – Brian M. Scott Nov 30 '12 at 22:45
  • half-open interval, right? how did you come to $\frac{3}{4}$? – doniyor Nov 30 '12 at 22:48
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    @doniyor: Yes, it’s a half-open interval, but which one? $\frac34\in[0,1)\subseteq C$, so $\frac34\in C$; I picked it more or less at random. – Brian M. Scott Nov 30 '12 at 22:50
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    @doniyor: Further hint: Do the intervals $\left[0,\frac1n\right)$ with $n>1$ actually contribute anything to the set $C$? – Brian M. Scott Nov 30 '12 at 22:57
  • @doniyor: No problem! – Brian M. Scott Dec 01 '12 at 12:09
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Since all of them are subsets of one of them, their union is just that one, namely $[0,1)$.

Michael Hardy
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