Question about isolated points.

  • Thread starter cragar
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  • #1
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Homework Statement


If I just had the set containing [itex] \pi [/itex] on the real line.
So this is an isolated point. Is this set closed?

The Attempt at a Solution


I think this set is closed because it contains its limit points, because it only has one point.
Am i thinking about this correctly?
 

Answers and Replies

  • #2
Hmm, that's an interesting question. I guess it would be closed because that's the only point in the set, and there's no option for an e-neighbourhood around it.
 
  • #3
Dick
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I think you are thinking about it correctly but that you are even asking this question makes me wonder. Can you explain your doubts further? What's the exact definition of a limit point?
 
  • #4
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thats what I thought too
 
  • #5
If we wanted to put things on more definite footing, we could assume that it isn't closed. Then there is some point x1 such that for all E > 0 there is some x in our set such that d(x,x1) < E. However, since x = π in all cases, then the only such point is π, a contradiction.
 

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