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

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 ?)

Numbers are the essence of the Universe

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

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

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

Numbers are the essence of the Universe

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