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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?

3 Answers 3

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Since all of them are subsets of one of them, their union is just that one, namely $[0,1)$.

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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.

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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 ?2012-12-01
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HINT: $[0,1)\supseteq\left[0,\frac12\right)\supseteq\left[0,\frac13\right)\supseteq\left[0,\frac14\right)\supseteq\ldots$

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    @doniyor: No problem!2012-12-01