An infinite union of closed sets?

In summary, an infinite union of closed sets can result in a set that is either open, closed, both, or neither, depending on the specific sets and their intersections. In the case of Cn= [-n, n], the union is the set of all real numbers which is both open and closed because of the definition of topology. However, in other cases, the infinite union may not have a clear characterization as open or closed.
  • #1
pivoxa15
2,255
1
Is this infinite union closed, open or neither? What is your reason?
 
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  • #2
Well, if C0= [0, 1] and Cn= a closed subset of [0,1] for all positive integers n, then each Cn is closed and, obviously, the union is C0= [0, 1], a closed set.

On the other hand, take a look at Cn= [1/n, 1-1/n] (intervals of real numbers) for n any positive integer. Cn is closed for all n, but the union of that collection of closed sets is (0, 1) which is open.

Or, let Cn= [1/n, 1]. Again, each Cn is closed but the union now is (0, 1] which is neither open nor closed.

But you left out a possibility: How about "both"? Let Cn= [-n, n] for all positive integers n. Each set is closed and the union is the set of all real numbers which is both open and closed.

In other words, just like the infinite intersection of open sets, nothing can be said about an infinite union of closed sets!
 
  • #3
HallsofIvy said:
But you left out a possibility: How about "both"? Let Cn= [-n, n] for all positive integers n. Each set is closed and the union is the set of all real numbers which is both open and closed.

In other words, just like the infinite intersection of open sets, nothing can be said about an infinite union of closed sets!



Do you mean the union is the set of all postive integers since you allowed n to be a positive integer?

You say that the union of this set is both closed and open. I can see how it is open because given any set in this infinite union, there will always be a neighbourhood of this union (in the form of a larger set) that is part of this union. But how is it closed? To be closed, you have to show that the complement of this infinite union is open.

I follow all your other examples. You seemed to be have stated different conclusions so it depends on the situation how an infinite union of closed sets will turn out.
 
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  • #4
pivoxa15 said:
Do you mean the union is the set of all postive integers since you allowed n to be a positive integer?
No each of thos sets is an interval, the union is R, exactly like he said.

You say that the union of this set is both closed and open. I can see how it is open because given any set in this infinite union, there will always be a neighbourhood of this union (in the form of a larger set) that is part of this union. But how is it closed? To be closed, you have to show that the complement of this infinite union is open.
The union is R, the complement is empty.

Several new problems now come to light though:

1. Neighbourhood of this union. What does that mean?

2. The set of integers Z in R which you claim is the union is certainly not open like you claim and it is in fact rather obviously closed which you claim not to see. Those are disturbing things. So what did you mean to say, precisely?
 
  • #5
matt grime said:
No each of thos sets is an interval, the union is R, exactly like he said.

I get this point now. I will continue to use this set as our example.
matt grime said:
The union is R, the complement is empty.

Several new problems now come to light though:

1. Neighbourhood of this union. What does that mean?

2. The set of integers Z in R which you claim is the union is certainly not open like you claim and it is in fact rather obviously closed which you claim not to see. Those are disturbing things. So what did you mean to say, precisely?
I used the word neighbourhood too loosely here. I meant whenever you have a set such as [-n,n] than there will always be another set whether larger such as [-n-1, n+1] or smaller such as [-n+1, n-1] which will lie in this infinite union. Hence the infinite union is open.

The complement of this set is the empty set as you point out. It is both open and closed according to http://en.wikipedia.org/wiki/Empty_set hence our infinite union could also be closed when taking the complement, the empty set as open.
 
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  • #6
No, I suggest you need to read the definition of open more closely.

Openness is a statement involving points and open sets containing points, nothing to do with closed sets containing other closed sets.The empty set regarded as a subset of a topological space is open and closed by the definition of topology.
 
  • #7
matt grime said:
No, I suggest you need to read the definition of open more closely.

Openness is a statement involving points and open sets containing points, nothing to do with closed sets containing other closed sets.


The empty set regarded as a subset of a topological space is open and closed by the definition of topology.


So the reason why 'Cn= [-n, n] for all positive integers n. Each set is closed and the union is the set of all real numbers which is both open and closed.' is because the complement of this set which is the empty set is both open and closed hence Cn is both closed and open.
 
  • #8
No, the reason R as a subset of R is open and closed is *because it is*: it satisfies the definition of being an open subset and the definition of being a closed subset.

X is Open <=> every point is contained in some open neighbourhood N contained in X
X is closed <=> Insert your favourite definition here.
 
  • #9
matt grime said:
No, the reason R as a subset of R is open and closed is *because it is*: it satisfies the definition of being an open subset and the definition of being a closed subset.

X is Open <=> every point is contained in some open neighbourhood N contained in X
X is closed <=> Insert your favourite definition here.


So in this case every point in R is contained in some open neighbourhood N ranging from -infinity to infinity. Therefore R as a subset of R is open.

R as a subset of R is closed because the complement of it is open.
 
  • #10
Not every one uses "its complement is open" as the definition of closed.
It is also possible to define "closed" as "contains all of its limit points".
Of course, if {an} is a convergent sequence of real numbers then its limit is a real number so R is closed under that definition.

Yet another definition of closed is "contains all of its boundary points" where a "boundary point" is a point such that every neighborhood contains some points in the set and some not in the set. (One can also then say that an open set "contains none of its boundary points). R itself has NO boundary points so it is correct to say that it contains none of its boundary points and that it contains all of its boundary points so it is both open and closed.
 
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  • #11
HallsofIvy said:
Not every one uses "its complement is open" as the definition of closed.
It is also possible to define "closed" as "contains all of its limit points".
Of course, if {an} is a convergent sequence of real numbers then its limit is a real number so R is closed under that definition.

Yet another definition of closed is "contains all of its boundary points" where a "boundary point" is a point such that every neighborhood contains some points in the set and some not in the set. (One can also then say that an open set "contains none of its boundary points). R itself has NO boundary points so it is correct to say that it contains none of its boundary points and that it contains all of its boundary points so it is both open and closed.

It worries me somewhat that you could make statements about something that does not exist. For example I am not a bird but I claim that I have a long beak and a short beak. It would be nonsensical for me to speak about beaks in the first place. But this is more philosophy than maths. It might be better to just define the set R as both open and closed, just like the empty set without digging for logical foundations.
 
  • #12
What does or doesn't exist? The empty set exists just as much as the set of real numbers.

Or perhaps you need to understand that logical constructs work in certain ways.

For instance for you to demonstrate that R does not contain all its boundary points you would have to produce one not in R. But as there is none, the statement

there exists a boundary point of R not in R

is false, hence

all boundary points of R are in R

is true.
 
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  • #13
matt grime said:
What does or doesn't exist? The empty set exists just as much as the set of real numbers.

I was trying to say that the fact that R has no boundary points would mean making agruments about them nonsensical.

In other words, the boundary points of R does not exist so we shouldn't mention it. We are only allowed to say that they don't exist and nothing else (i.e. can't have a statement starting with "supose R contains all of its boundary points...")


matt grime said:
Or perhaps you need to understand that logical constructs work in certain ways.

For instance for you to demonstrate that R does not contain all its boundary points you would have to produce one not in R. But as there is none, the statement

there exists a boundary point of R not in R

is false, hence

all boundary points of R are in R

is true.

I understand what you are getting at here.

Prove: R contains all of its boundary points
1) Try to show a boundary point of R not in R.
2) 1 can't happen since R has no boundary points
Hence R is closed

Prove: R contains none of its boundary points
1) Try to show a boundary point of R in R
2) 1 can't happen since R has no boundary points
Hence is R is open

The possible flaw I see in these proofs is that the statement trying to be proved is nonsensical since 'R contains ... its boundary points.' is impossible. So not only step 1 (in each statement) can't happen but what we are trying to prove cannot happen either.
 
  • #14
HallsofIvy said:
Not every one uses "its complement is open" as the definition of closed.
It is also possible to define "closed" as "contains all of its limit points".
Of course, if {an} is a convergent sequence of real numbers then its limit is a real number so R is closed under that definition.

Yet another definition of closed is "contains all of its boundary points" where a "boundary point" is a point such that every neighborhood contains some points in the set and some not in the set. (One can also then say that an open set "contains none of its boundary points). R itself has NO boundary points so it is correct to say that it contains none of its boundary points and that it contains all of its boundary points so it is both open and closed.
I have just realized something 'illogical' about the definition of open sets according to 'R contains none of its boundary points'. Surely 'its boundary points' refer to R's boundary points so R has boundary points. But than to state R contains none of its boundary points is confusing.
 
  • #15
Surely 'its boundary points' refer to R's boundary points so R has boundary points.
Yep. And its set of boundary points is the empty set.
 
  • #16
Since the definition of open is not 'does not contain any of its boundary points' I'im not going to worry unduly about this.

However the statement 'if x is in the boundary of R then X' is always going to be true since the hypothesis is false; 'x in the boundary of R is always' false hence it implies anything you want. it's just elementary (in the non-perjorative use of the word logic).
 
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  • #17
matt grime said:
Since the definition of open is not 'does not contain any of its boundary points' I'im not going to worry unduly about this.

The definition of open set can vary from textbook to textbook. I have certainly seen texts on metric topology where they introduce the notion of a neighborhood [itex]N_\delta(x)= \{y|d(x,y)< \delta\}[/itex], then define "p is an interior point of set A" as "there exist some [itex]\delta[/itex] such that [itex]N_\delta(x)[/itex] is a subset of A", define "p is an exterior point of set A" as "p is an interior point of the complement of A", and "p is a boundary point of A" as "p is neither an interior point nor an exterior point of A". Then the definition of "open set" is "contains none of its boundary points" and "closed set" is "contains all of its boundary points".

Since all points in the universal set fall into one and only one of "interior point", "exterior point", "boundary point", and obviously exterior points of A can't be in A, saying that A contains none of its boundary points is exactly the same as saying all of its points are interior points. Since A and its complement have exactly the same boundary points, saying A contains all of its boundary points is exactly the same as saying its complement is open.

My experience is that students who have been dealing with open and closed intervals since before calculus typically accept those definitions more easily than "all of its points are interior points" for open and "its complement is open" for closed.
 
  • #18
That was badly worded on my part. If I were to say anything then I should have said: that is not what the definition of open should be.
 
  • #19
HallsofIvy said:
The definition of open set can vary from textbook to textbook. I have certainly seen texts on metric topology where they introduce the notion of a neighborhood [itex]N_\delta(x)= \{y|d(x,y)< \delta\}[/itex], then define "p is an interior point of set A" as "there exist some [itex]\delta[/itex] such that [itex]N_\delta(x)[/itex] is a subset of A", define "p is an exterior point of set A" as "p is an interior point of the complement of A", and "p is a boundary point of A" as "p is neither an interior point nor an exterior point of A". Then the definition of "open set" is "contains none of its boundary points" and "closed set" is "contains all of its boundary points".

Since all points in the universal set fall into one and only one of "interior point", "exterior point", "boundary point", and obviously exterior points of A can't be in A, saying that A contains none of its boundary points is exactly the same as saying all of its points are interior points. Since A and its complement have exactly the same boundary points, saying A contains all of its boundary points is exactly the same as saying its complement is open.

My experience is that students who have been dealing with open and closed intervals since before calculus typically accept those definitions more easily than "all of its points are interior points" for open and "its complement is open" for closed.
I am much more comfortable with "all of its points are interior points" for open and "its complement is open" for closed but than again I have done some calculus and also because they are the exact definitions I have been taught to learn.

With your prior definitions. Do you have to stick points P to the boundary of a set S and see whether there are points in S which exactly match the points P? If they do than S is closed. If they don't than S is open.
 
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  • #20
matt grime said:
Since the definition of open is not 'does not contain any of its boundary points' I'im not going to worry unduly about this.

However the statement 'if x is in the boundary of R then X' is always going to be true since the hypothesis is false; 'x in the boundary of R is always' false hence it implies anything you want. it's just elementary (in the non-perjorative use of the word logic).

So you are implying the material conditional in loigc? For 'if a than b' to be true and when a is false, b can be either true or false. That makes a bit more sense.
 
  • #21
Can you give me hints, please, about how to proof that an infinite collection of closed neighborhoods in RxR is also closed?

I am aware that such it is not necessarily the case in R (as Hallsoftivy has shown upwards). But in some proof Apostol uses that statement (leaving the proof to the reader).
 
  • #22
Castilla said:
Can you give me hints, please, about how to proof that an infinite collection of closed neighborhoods in RxR is also closed?

I am aware that such it is not necessarily the case in R (as Hallsoftivy has shown upwards). But in some proof Apostol uses that statement (leaving the proof to the reader).
R is a closed subspace of RxR, and so any counterexample in the former will also be a counterexample in the latter...
 
  • #23
Okey. An infinite union of closed sets in RxR no necessarily is closed... but, in the particular case of an infinite union of closed discs in RxR, it is closed or it is not?
 
  • #24
Castilla said:
Okey. An infinite union of closed sets in RxR no necessarily is closed... but, in the particular case of an infinite union of closed discs in RxR, it is closed or it is not?
Definitely not. It's easy enough to construct a nested collection of discs whose union is an open disc. Also, consider the simplest (in some sense) possible counterexample to the previous claim: a sequence of points converging to some point not in the sequence. It's easy enough to tweak this example by replacing each point with a closed disc of positive radius. (and have them be nonoverlapping!)
 
  • #25
Hurkyl, thanks for your insights. I was wondering because in another thread I find this question:

Problem: given compact set C and an open set U, with C subset of U, show there is a compact set D, subset of U, with C being interior of D.

And Mathwonk answered: "(asuming an euclidian space) if so there is a minimum distance of points of your compact set from the closed exterior of your open set, so cover your compact set with little closed balls of radius half that distance."

Mathwonks is implying that the union of little closed balls or radius half said distance is a compact set (therefore a closed set).

There must be something I am missing.
 
  • #26
Castilla said:
Hurkyl, thanks for your insights. I was wondering because in another thread I find this question:

Problem: given compact set C and an open set U, with C subset of U, show there is a compact set D, subset of U, with C being interior of D.

And Mathwonk answered: "(asuming an euclidian space) if so there is a minimum distance of points of your compact set from the closed exterior of your open set, so cover your compact set with little closed balls of radius half that distance."

Mathwonks is implying that the union of little closed balls or radius half said distance is a compact set (therefore a closed set).

There must be something I am missing.
Mathwonk never said to consider the union of all of those balls... While that union is compact, there is a much simpler approach.

But if you really want to do it that way... it's fairly straightforward to show that union contains all of its boundary points, and thus closed. (It's also easy to show it's bounded, and thus also compact) It might help to try and come up with an alternate description of that set...
 
  • #27
"But if you really want to do it that way... it's fairly straightforward to show that union contains all of its boundary points, and thus closed. (It's also easy to show it's bounded, and thus also compact) It might help to try and come up with an alternate description of that set... "

Please let out some hint...
 
  • #28
Castilla said:
"But if you really want to do it that way... it's fairly straightforward to show that union contains all of its boundary points, and thus closed. (It's also easy to show it's bounded, and thus also compact) It might help to try and come up with an alternate description of that set... "

Please let out some hint...
Can you come up with a good criterion to tell whether or not a particular point is in that set?
 
  • #29
We have the curve K inside an open region S. We cover K with open discs centered in each point z of K and all of them with the same fixed radius "r" which allows all of them to be inside of S.

Now we take the closed discs N(z, r/2). We have to prove that this last set of discs (which I call "T") is compact. I know it is bounded (it is inside S). To prove closedness I tried this:

A closed set contains all its accumulation points. Let's suppose T does not. So there it is a point "a" which is a.p. of T and does not belong to T. Therefore (remember T is inside S) "a" belongs to the interior of S - T.
So there is a disc N(a) in the interior of S - T.
But, being "a" an a.p. of T, there are points of T inside N(a)
But N(a) is inside of S - T. So we got points of T inside S - T. Contradiction.

But then I realized my work was flawed, because of this deduction: "So there it is a point "a" which is a.p. of T and does not belong to T. Therefore "a" belongs to the interior of S - T."

That deduction is not true, because "a" no necesarily is in the interior of S - T. The point "a" might be located in the boundary of T, so I can not take a disc N(a) to follow up my "proof".

How can I fix this?
Thanks fr your help.

(I found a similar thread in the Algebra forum so I am posting the same SOS message there).
 
  • #30
I think I have seen something more.

In my last post I saw that my "proof" was flawed because "a" might be in the boundary of T. (I will denote it B(T) ).

But a boundary is always a closed set. Now let's take one of the compact (closed and bounded) discs of T, denoting it "D". We are supposing that B(T) and disc D do not intersect. (If they did it, the problem would be solutioned). Then there is a distance between B(t) and the disc D.

So if "a" is in B(T) , there is a fixed distance between it and its "accumulated points" which all belong to one or other closed disc. End.

Please tell me if this is ok.
 
  • #31
Er, Hurkyl, maybe you can tell me if that was Ok?
 
  • #32
pivoxa15 said:
I have just realized something 'illogical' about the definition of open sets according to 'R contains none of its boundary points'. Surely 'its boundary points' refer to R's boundary points so R has boundary points. But than to state R contains none of its boundary points is confusing.
You do understand, I hope, that "A is closed if it contains all of its boundary points" is not intended to be a strict, logical definition. It is shorthand for "If the statement (if x is a boundary point of A the x is contained in A) is true then A is closed". For A= R, the statement "if x is a boundary point of R then x is contained in R" is true for all R because the premises are false.

If every statement had to be in rigorously true logic, then every mathematics book would be as long (and as hard to read) as "Principia Mathematica"!
 

1. What is an infinite union of closed sets?

An infinite union of closed sets is a mathematical concept where an infinite number of closed sets are combined together. A closed set is a set that contains all of its limit points, meaning that it includes its boundary points. The union of these sets is a new set that contains all the elements from each individual set.

2. How is an infinite union of closed sets different from a finite union?

An infinite union of closed sets is different from a finite union in that it includes an infinite number of sets, while a finite union only includes a finite number of sets. Additionally, the infinite union may contain elements that are not in any of the individual sets, while the finite union only contains elements that are in at least one of the sets.

3. What is the purpose of studying infinite unions of closed sets?

The study of infinite unions of closed sets is important in mathematics, particularly in topology and analysis. It allows us to understand the properties and behavior of infinite sets, and can be applied to various real-world problems such as optimization and approximation.

4. How is an infinite union of closed sets related to the concept of closure?

An infinite union of closed sets is closely related to the concept of closure. The closure of a set is the smallest closed set that contains all the points of the original set. In other words, it is the union of the original set and all of its limit points. Therefore, an infinite union of closed sets can be seen as a way to construct the closure of a set.

5. Can an infinite union of closed sets ever be a closed set itself?

Yes, an infinite union of closed sets can be a closed set itself. This can happen if the sets in the union are nested, meaning that each set is a subset of the next one. In this case, the infinite union will also be a closed set, as it will contain all of its limit points.

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