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Topology question; derived pts and closure

  • Thread starter Unassuming
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  • #1
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Homework Statement


If A is a discrete subset of the reals,

prove that

[tex]A'=cl_x A \backslash A[/tex]

is a closed set.


Homework Equations


A' = the derived set of A
x is a derived pt of A if [tex]U \cap (A \backslash \{x\}) \neq \emptyset[/tex] for every open U such that x is in U.

Thrm1. A is closed iff A=cl(A)
Thrm2. cl(emptyset)=emptyset


The Attempt at a Solution



"Proof". Using Thrm1 with A' we can see that A' is closed iff A'=cl(A'). Since A is a discrete subset of the Reals we know that the set consists of isolated pts. Since there are no derived pts in A, then A'= emptyset. Using Thrm2 we know that cl(emptyset)=emptyset. Therefore A'=cl(A'). And thus A' is closed.

Does this proof work?
 

Answers and Replies

  • #2
Dick
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I think so. Is there any difference between cl_x(A) and cl(A)? A' is closed because it's empty.
 
  • #3
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Well, a classmate pointed out to me that the set B={1/n : n in Naturals}, has a derived point of {0}. I feel like this set is discrete, yet it's derived points are not the empty set.

I thought I had this one. Any ideas of where I have gone wrong? My guess would be assuming that discrete subsets have no derived points but then what direction do I take?
 
  • #4
Dick
Science Advisor
Homework Helper
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Whoa. You're right. You caught me. I thought the proof sounded a little vacuous. So you want to prove that if x is not in A', then there is a neighborhood of x that does not intersect A'. If x is in A, use that A is discrete. If x is not in A, then use that A is closed. Can you take it from there?
 

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