Every element of a set [itex]A[/itex] can be written [itex]a=w.a_1a_2a_3\ldots{a_n}\ldots[/itex] with [itex]w, a_n\in\mathbb{Z}[/itex] and [itex]0\leq a_n\leq9[/itex] for every [itex]n\in\mathbb{N}.[/itex] If [itex]A[/itex] is bounded, there exists a greatest whole part [itex]\overline{w}[/itex] of the elements of [itex]A,[/itex] and because any set [itex]S[/itex] of elements [itex]a_n[/itex] is bounded, for every [itex]n,[/itex] there exists a greatest element [itex]\overline{a_n}[/itex] of [itex]S.[/itex](adsbygoogle = window.adsbygoogle || []).push({});

Let [itex]A_0\subset{A}[/itex] be the set of elements [itex]a=\overline{w}.a_1a_2a_3\ldots{a_n}\ldots[/itex] and [itex]A_1\subset{A_0}[/itex] the set of elements [itex]a=\overline{w}.\overline{a_1}a_2a_3\ldots{a_n}\ldots,[/itex] and define [itex]A_{n+1}\subset{A_n}[/itex] as the set of elements [itex]a=\overline{w}.\overline{a_1}\overline{a_2}\overline{a_3}\ldots\overline{a_n}\overline{a_{n+1}}a_{n+2}\ldots.[/itex]

It would appear as though [itex]\alpha\in\bigcap_{n=1}^\infty{A_n}[/itex] contains a single element--namely, [itex]\sup(A)[/itex]. However, this seems to lead to the absurd conclusion that for every set [itex]A,\sup(A)\in{A}[/itex]. Perhaps we would say simply that [itex]\bigcap_{n=1}^\infty{A_n}=\emptyset[/itex] in case [itex]\sup(A)\not\in{A}[/itex]?

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# Infinite intersection of indexed sets

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