ONT Re: Topology
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| 1. Topological Spaces
|
| 1.7. Relativization, Separation (cont.)
|
| If (Y, !U!) is a subspace of (X, !T!) and Y is open in X,
| then each set open in Y is also open in X because it is the
| intersection of an open set and Y. A similar statement, with
| "closed" replacing "open" everywhere, is also true. However,
| knowing that a set is open or closed in a subspace generally
| tells very little about the situation of the set in X. If X
| is the union of two sets Y and Z and if A is a subset of X
| such that A |^| Y is open in Y and A |^| Z is open in Z,
| then one might hope that A is open in X. But this is
| not always true, for if Y is an arbitrary subset of X
| and Z = X ~ Y, then Y |^| Y and Y |^| Z are open in Y
| and Z respectively. There is one important case,
| in which this result does hold.
|
| Two subsets A and B are 'separated' in a topological space X
| iff A˜ |^| B and A |^| B˜ are both void. This definition of
| separation involves the closure operation in X. However, the
| apparent dependence on the space X is illusory, for A and B are
| separated in X if and only if neither A nor B contains a point or
| an accumulation point of the other. This condition may be restated
| in terms of the relative topology for A |_| B, in view of part (b)
| of the foregoing theorem, as: Both A and B are closed in A |_| B
| (or, equivalently, A (or B) is both open and closed in A |_| B) and
| A and B are disjoint. As an example, notice that the open intervals
| (0, 1) and (1, 2) are separated subsets of the real numbers with the
| usual topology and that there is a point, 1, belonging to the closure
| of both. However, (0, 1) is not separated from the closed interval
| [1, 2] because 1, which is a member of [1, 2], is an accumulation
| point of (0, 1).
|
| JLK, Gen Top, page 52.
|
| John L. Kelley, 'General Topology',
| Van Nostrand Reinhold, New York, NY, 1955.
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