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Limit points

  • #1

Homework Statement



Let X be a metric space and let E be a subset of X. Prove that E' is closed and that E and E-closure have the same set of limit points.

Homework Equations





The Attempt at a Solution



I have proven that E' is closed. Now, from the definition of a closed subset of a metric space, I know that if E' is closed then every point of E' is a limit point. I also know that E-closure = E ∪ E'. Then, is it right to conclude that (E-closure)' = E' ∪ (E')' = E' ?
 

Answers and Replies

  • #2
Dick
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I'm assuming E' means the boundary of E. Is it? You'll probably want to state your definitions of all of these terms. They aren't always the same in all texts. How did you prove E' is closed?
 
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  • #3
I'm assuming E' means the boundary of E. Is it? How did you prove E' is closed?
E' is the set of limit points of E.
Here is how I proved it,

Let [itex]X[/itex] be a metric space and let [itex]E\subset X[/itex]. Let [itex]p\in X-E'[/itex]. Since [itex]p[/itex] is not a limit point of [itex]E[/itex], there exists [itex]\epsilon >0[/itex] so that [itex]N_{\epsilon}(p)\cap E = \emptyset[/itex].
Now, let [itex]a\in E'[/itex] and [itex]b\in E[/itex]. Then, there exists [itex]\delta >0[/itex] so that [itex]|a-b|<\delta[/itex].
Let [itex]\delta = \frac{\epsilon}{2}[/itex]. Then, [itex]|p-a| = |p-b+b-a|
\geq ||p-b|-|a-b||[/itex]. But [itex]\epsilon=2\delta[/itex]. Then, [itex]|p-a| > \epsilon - \delta = 2\delta - \delta = \delta[/itex]. So [itex]|p-a| > \delta[/itex]. Then,
[itex]N_{\delta}(p) \cap E' = \emptyset[/itex]. Then, [itex]N_{\delta}(p) \subset X-E'[/itex]. Then, every [itex]p\in X-E'[/itex] is an interior point. So [itex]X-E'[/itex]is open. Therefore, [itex]E'[/itex] is closed. q.e.d

The Nε(p) represents, for example, an epsilon neighborhood of p.
 
  • #4
Dick
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Without considering all the details, that sounds generally right. How do define 'limit point'? If p is an element of E, is it a limit point? Sorry to ask all these questions.
 
  • #5
Hey :smile:. This is the definition from Rudin,

Let X be a metric space and let E be a subset of X. A point p in X is a limit point of E if every neighborhood of p contains a point q different from p such that q is in E.
 
  • #6
Dick
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Hey :smile:. This is the definition from Rudin,

Let X be a metric space and let E be a subset of X. A point p in X is a limit point of E if every neighborhood of p contains a point q different from p such that q is in E.
Ok, then it's not generally true that (AUB)'=A'UB'. Pick A={0} and B={1/n for n>=1}. Then (AUB)'={0} but A'={} and B'={}. You'll have to come up with a better argument.
 
  • #7
Ok, then it's not generally true that (AUB)'=A'UB'. Pick A={0} and B={1/n for n>=1}. Then (AUB)'={0} but A'={} and B'={}. You'll have to come up with a better argument.
But if you use the archimedean property of real numbers, then B' = {0} right? So in that case (AUB)' = A'UB' :confused:.
 
  • #8
Dick
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But if you use the archimedean property of real numbers, then B' = {0} right? So in that case (AUB)' = A'UB' :confused:.
Of course that's true. The confusion is all mine. p is a limit point of AUB iff p is a limit point of A or p is a limit point of B. So (AUB)'=A'UB' in general. Something was bothering me last night and now I don't know what it was. I agree with your argument.
 

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