x epsilon1 s x But since x S x epsilon1 s x From this we conclude that every

X epsilon1 s x but since x s x epsilon1 s x from this

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x - epsilon1 < s x . But, since x / S , x - epsilon1 < s < x . From this we conclude that every deleted neighborhood N * ( x, epsilon1 ) contains a point of S , so x is an accumulation point of S . 15 Let x be an accumulation point of S . Let N ( x, epsilon1 ) be a neighborhood of x . Since x is an accumulation point of S , the deleted neighborhood N * ( x, epsilon1 ) contains a point s 1 S ; the deleted neighborhood N * ( x, epsilon1/ 2) contains a point s 2 S ; the deleted neighborhood N * ( x, epsilon1/ 3) contains a point s 3 S , and so on. The set { s 1 , s 2 , s 3 , · · ·} ⊆ N ( x, epsilon1 ) is an infinite subset of S . 2
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  • Fall '08
  • Staff
  • Math, Empty set, Order theory, β, α, Closed set, General topology

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