Confused with closure and interior

In summary: So the closure will always contain A. You can argue that 1 is a limit point of A, hence must be an element of the closure.
  • #1
Greetings all,

I'm looking at some examples in the Topology: Pure and Applied text.

Looking at example 2.1 Consider A=[0,1) as a subset of R with the standard topology. Then Aint=(0,1) and Aclos=[0,1].

Can someone explain to me why the union of all open sets in A is that?

Furthermore, example 2.3

Consider A=[0,1) as a subset of R in the finite complement topology. Here Aint = 0 because there are no nonempty open sets contained in [0,1). Since A is infinite, and the only infinite closed set in this topology is R, it follows that Aclos=R.

I have the same question as above, but also,why is A infinite? And more importantly, the book described R as being open when proving the the finite complement is a topology, and now they are saying its closed?

Please help!
 
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  • #2
Damascus Road said:
Greetings all,

I'm looking at some examples in the Topology: Pure and Applied text.

Looking at example 2.1 Consider A=[0,1) as a subset of R with the standard topology. Then Aint=(0,1) and Aclos=[0,1].

Can someone explain to me why the union of all open sets in A is that?
Is which? The union of all open sets in A, (0, 1), is the interior of A. It is trivial that there is no larger open subset of A, as A - (0, 1) is a singleton. Constructively, take the union of all disc neighborhoods of points in A whose intersection with R - A is null. The closure of A is the set of all points for which every neighborhood has an intersection with A that is not the null set.

Furthermore, example 2.3

Consider A=[0,1) as a subset of R in the finite complement topology. Here Aint = 0 because there are no nonempty open sets contained in [0,1). Since A is infinite, and the only infinite closed set in this topology is R, it follows that Aclos=R.

I have the same question as above, but also,why is A infinite? And more importantly, the book described R as being open when proving the the finite complement is a topology, and now they are saying its closed?
The entire space is always both open and closed in any topology. This is a consequence of the definition of open and closed subsets of R in combination with the null set being a subset of every set, and is stated explicitly in the definition of a topological space.
The finite complement topology uses the amount of elements of a set to define open or closed; there are an infinite amount of elements in any real interval, which is why A is described as being infinite, and therefore open.
 
  • #3
Thank you for the reply, I had forgotten about that property of being both open and closed.

You're explanation has helped a lot!

The thing I am still struggling with is the closure now, the "intersection of all closed sets in A".

First, could I not choose a "set" that is closed in A that doesn't include 0 or 1?
Then the intersection would be empty...

Second, if the set A is open on 1 (i.e. doesn't reach one) how can it be included in the closure?
 
  • #4
Damascus Road said:
The thing I am still struggling with is the closure now, the "intersection of all closed sets in A".

The closure is the intersection of all closed sets containing A. So the closure will always contain A. You can argue that

1) [0,1] is a closed set containing A.
2) 1 is a limit point of A, hence must be an element of the closure.
 
  • #5
Damascus Road said:
The thing I am still struggling with is the closure now, the "intersection of all closed sets in A".

First, could I not choose a "set" that is closed in A that doesn't include 0 or 1?
Then the intersection would be empty...
You can't choose sets in any manner, you must intersect all closed sets that contain A.

Second, if the set A is open on 1 (i.e. doesn't reach one) how can it be included in the closure?
A closed set C containing A must contain 1. If it did not contain 1, then the boundary of the closed set must necessarily include a number r less than 1. Under the standard topology, you can simply choose a disc neighborhood around that number with radius 1-r that includes points not in C, but in A, contradicting the requirement that C contains A.
 

1. What is the difference between closure and interior?

Closure refers to the set of all points that can be reached from a given point by following a sequence of limit points. Interior, on the other hand, refers to the set of all points that lie within a given set and do not touch the boundary.

2. How are closure and interior related?

The closure of a set includes the set's interior, as well as its boundary. In other words, the closure contains all the points in the interior, as well as all the points on the boundary. However, the interior does not include any points on the boundary.

3. Can a set have both an empty closure and an empty interior?

Yes, a set can have both an empty closure and an empty interior. This occurs when the set itself is empty. Since there are no points in the set, there are also no limit points or points within the set, resulting in an empty closure and interior.

4. How do closure and interior relate to open and closed sets?

A set is open if all its points are in the interior, and it is closed if it contains all its boundary points. Therefore, the interior of a set is open, and the closure is closed. In general, a set can be both open and closed, or neither.

5. Why is understanding closure and interior important in mathematics?

Understanding closure and interior is crucial for many areas of mathematics, including topology, real analysis, and measure theory. These concepts help us define and characterize different types of sets, which are fundamental building blocks in mathematical structures and theories.

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