R(=real Nos)

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  • #1
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Main Question or Discussion Point

Given that f is a function from R(=real Nos) to R continuous on R AND ,A any subset of R,IS THE closure of f(A) ,a closed set??
 

Answers and Replies

  • #2
mathman
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The closure of any set is by definition a closed set. I think you should rephrase your question.
 
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  • #3
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Yes, you right thank you. But if we define a set to be closed if its complement is open,
how then we prove its closure to be a closed set??
 
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  • #4
HallsofIvy
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What definition of "closure of A" are you using?
 
  • #5
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Yes, you right thank you. But if we define a set to be closed if its complement is open,
how then we prove its closure to be a closed set??
what? by definition the closure of a set A is the smallest closed set that contains A.
 
  • #6
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I presume the OP had in the mind the definition that the closure of S is the union of S and the set of its limit points. In this case:

Denote by S' the closure of S. Then we wish to show that S' is closed. Suppose x is in the complement of S'. Then x is not in S and is not a limit point of S. So there is an open ball around x that doesn't intersect S. This open ball cannot contain any limit point of S since if y is inside it, then there is a smaller ball centered at y contained in the bigger - and so there is an open ball around y that doesn't intersect S, so y is not a limit point of S. It follows that the open ball around x does not intersect S'. Therefore the complement of S' is open; so S' is closed.
 

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