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Topology - Interior of set - Rudin

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Homework Statement



I am trying to solve part d of problem 9 in chapter 2 of Rudin's Principles of Mathematical Analysis. The problem is: Let E* denote the set of all interior points of a set E (in a metric space X). Prove the complement of E* is the closure of the complement of E.

I will use E^c to denote the complement of E. It might be helpful to pull out Rudin and follow along; my notation is poor and the problem will make more sense.



2. The attempt at a solution
Proof:
First, we show that the complement of the interior points of E is a subset of the closure of the complement of E. Because the interior of E lies in E, the complement of E lies in the complement of the interior of E. Then because the complement of the interior of E is closed, the closure of the complement of E lies in the complement of the interior of E.

Now we need to show (E*)^c is in the closure of E^c. Pick an arbitrary point p in (E*)^c, Then it is not an interior point of E, and all neighborhoods around it contain a point of E^c. Then p is a limit point of E^c, and lies in the closure of E^c.


3. The error!
I have just shown that (E*)^c= closure(E^c). Now E*^c is obviously closed, because the interior of E is always open (this is part a of the problem). The second paragraph shows that any point in E*^c is a limit point of E^c. Now if E is open, E* and E are the same, so this would show that any point in E^c is a limit point of E^c. This would mean E^c is always a perfect set (closed and every point is a limit point). Surely this is much too strong: the complement of an arbitrary open set is not always perfect!?
 

Answers and Replies

  • #2
PAllen
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I'm not sure there is any error. It seems plausible to me that the complement of an open set in a metric space includes all its limit points and is closed. (I am not an expert in this, but I am, say, 95% sure this is correct).
 
  • #3
HallsofIvy
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PAllen, what you are saying it true- but irrelevant. The question is not whether it contains all of its limit points, which is true of any closed set, but whether all of its points are limit points.

Consider the set E= (-1, 0)U(0, 1). That is an open set. Its complement is (-inf, 1]U {0}U[1, inf) which is closed but not a perfect set because 0 is not a limit point.
 
  • #4
PAllen
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PAllen, what you are saying it true- but irrelevant. The question is not whether it contains all of its limit points, which is true of any closed set, but whether all of its points are limit points.

Consider the set E= (-1, 0)U(0, 1). That is an open set. Its complement is (-inf, 1]U {0}U[1, inf) which is closed but not a perfect set because 0 is not a limit point.
Yes, thanks, I see. So it is the second paragraph of the OP 'proof' that is flawed, making an unwarranted assumption. The desired result has to achieved a different way.
 
  • #5
HallsofIvy
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Yes. Specifically, the statement "Pick an arbitrary point p in (E*)^c, Then it is not an interior point of E, and all neighborhoods around it contain a point of E^c" is not true. In my example, 0 is a point of (E*)^c= E^c but no neighborhod around it contains another point in E^c. What is true is that any point in (E*)^c is either a limit point of E^c or is an isolated point. In either case it is in the closure of E^c.
 

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