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#1 2007-04-18 12:09:34

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

confuse

First , the book says, A component interval Sof S is not a proper subset of any other open interval contained in S .
Second , Every point of a nonempty open set S belongs to one and only one component interval of S.

What confuses me is that , the next theorem. Every nonempty open set S in R_1 is the union of a countable collection of disjoin component intervals of S.   (There is only one component interval in every S ! how come union  of ?)


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#2 2007-04-18 13:27:06

Zhylliolom
Real Member
Registered: 2005-09-05
Posts: 412

Re: confuse

Consider S = (1, 2) ∪ (3, 4), for example. Clearly (1, 2) is not a subset of (3, 4) and vice versa. So S has two component intervals, (1, 2) and (3, 4). Then a set S can have more than one component interval. Is this clear?

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#3 2007-04-18 15:17:05

Stanley_Marsh
Member
Registered: 2006-12-13
Posts: 345

Re: confuse

OH yeah , I always treat a set as an interval~ it can be serveral.


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