Finding the Interior and Closure of Sets in a Topological Space

[radou]

I'm not sure about my answers, any help is highly appreciated.

Let (N, U) be a topological space, where N is the set of natural numbers (without 0), and U = {0} U {Oi, i is from N}, where Oi = {i, i+1, i+2, ...} and {0} is the empty set. One has to find the interior (Int) and closure (Cl) of these sets:

(a) A = {n from N : (n - 7)/(n - 11) > 0}
(b) A = {13, 5, 2010}

So, for (a):

Obviously, A = {n from N : n > 11 or n < 7} = N \ {7, 8, ... , 11}. The interior of A, Int(A), is by definition the union of all open subsets contained in A, and the open subsets in the topology (N, U) are elements of the family U or their unions and finite intersections. So Int(A) = {0} (the empty set, since no open subset in U can contain A. Now, the closure of A, Cl(A), is by definition the intersection of all closed subsets which contain A, i.e. the smallest one of them. So, Cl(A) = N, and N is closed, since its complement is the empty set, which is open. (I feel I'm missing something huge here.)

(b) Again, Int(A) = {0}. What would Cl(A) be? The first guess is Cl(A) = {5, 6, ...}, but is this set closed? Its complement is {1, 2, 3, 4}, but this is no open set contained in U?

I hope I didn't cause much confusion, but I need to solve this problem in order to clear out my way of thinking.

 

[HallsofIvy]

Are you sure that "{0}" is the empty set? That's a very strange way to represent it. Normally, "{ }" is the empty set, and "{0}" is the set whose only member is "0".

 

[radou]

Oops, actually, I didn't find the symbol for the empty set, so I used "0" to represent it - so, we're talking about the set containing the empty set, {0} (this is exactly what's written in the problem I'm working on).

 

[slider142]

I'm not sure about my answers, any help is highly appreciated.

Let (N, U) be a topological space, where N is the set of natural numbers (without 0), and U = {0} U {Oi, i is from N}, where Oi = {i, i+1, i+2, ...} and {0} is the empty set. One has to find the interior (Int) and closure (Cl) of these sets:

(a) A = {n from N : (n - 7)/(n - 11) > 0}
(b) A = {13, 5, 2010}

So, for (a):

Obviously, A = {n from N : n > 11 or n < 7} = N \ {7, 8, ... , 11}. The interior of A, Int(A), is by definition the union of all open subsets contained in A, and the open subsets in the topology (N, U) are elements of the family U or their unions and finite intersections. So Int(A) = {0} (the empty set, since no open subset in U can contain A.

The set O₁₂ = {12, 13, ...} is contained in A and is also an element of U, and thus is open in (N, U). It is therefore contained in Int(A) as well.

 

[radou]

OK, so Int(A) is the union of the sets Oi, where i >= 12, right? What about the closure of A?

 

[slider142]

Are you using the concept of limit points, or the concept of the frontier of A: Fr(A) ?
In either case, what do the neighborhoods of points look like in (N, U)?

 

[radou]

Are you using the concept of limit points, or the concept of the frontier of A: Fr(A) ?
In either case, what do the neighborhoods of points look like in (N, U)?

Both of these concepts are defined later on in the exercise notes I'm going through, so I guess I don't need to use them.

If i is an element of N, then ... , Oi-1, Oi, Oi+1, ... are all its neighborhoods, since they're open sets in U containing i. Now, an element i is in the closure of A iff every neighborhood of x intersects A, so Cl(A) = {1, 2, ... , 11} (since for 12 and greater elements of N, one can find neighborhoods which don't intersect A).

 

[slider142]

Both of these concepts are defined later on in the exercise notes I'm going through, so I guess I don't need to use them.

If i is an element of N, then ... , Oi-1, Oi, Oi+1, ... are all its neighborhoods, since they're open sets in U containing i.

O_i+1 does not contain i. Only the sets O_n where n is less than or equal to i contain i.

Now, an element i is in the closure of A iff every neighborhood of x intersects A, so Cl(A) = {1, 2, ... , 11} (since for 12 and greater elements of N, one can find neighborhoods which don't intersect A).

Every neighborhood of 12 intersects A, since A contains all natural numbers greater than 11.

 

[radou]

Wow, I just realized what silly stuff I wrote down. Let's start over.

If x is an element of N, then every neighborhood of x intersects A, so CL(A) = N, right?

 

[slider142]

Wow, I just realized what silly stuff I wrote down. Let's start over.

If x is an element of N, then every neighborhood of x intersects A, so CL(A) = N, right?

Yep. :)

 

[radou]

OK, and for the set {13, 5, 2010}. Its closure equals N, and its interior is the empty set, right? (Since no open set from U is contained in this set) Thanks in advance.

 

[slider142]

OK, and for the set {13, 5, 2010}. Its closure equals N, and its interior is the empty set, right? (Since no open set from U is contained in this set) Thanks in advance.

This set's closure cannot be N, because the set O₂₀₁₁, an open set containing 2011, does not intersect it. The interior is correct.

 

[radou]

This set's closure cannot be N, because the set O₂₀₁₁, an open set containing 2011, does not intersect it. The interior is correct.

Oh yes, right. Thanks a lot. :)

So, the closure of A is {1, 2, ... 2010} or O1 \ O2011, right?

 

[slider142]

Oh yes, right. Thanks a lot. :)

So, the closure of A is {1, 2, ... 2010} or O1 \ O2011, right?

Yep, that's it. :)