Excluded Point Topology: Int(A) and Cl(A) for Sets A with or without p in X

In summary: For A to be open, it must be the case that there exists an open set containing A. For A to be closed, it must be the case that there exists a closed set containing A. A is open if and only if there exists an open set containing A. A is closed if and only if there exists a closed set containing A.
  • #1
cragar
2,552
3

Homework Statement


Consider the excluded point topology on a set X.
Determine Int(A) and Cl(A) for sets A containing p and for sets A not containing p.
Excluded point topology is all the subsets of X that exclude p.
where p is in X.

The Attempt at a Solution


So the interior of a set A is the union of all open sets contained in A.
Would the interior for A be A-p , where we exclude p.
and would the interior be A where we include p.
I am not sure what the smallest closed set would be that contains A.
It seem like it would just be A.
 
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  • #2
cragar said:

Homework Statement


Consider the excluded point topology on a set X.
Determine Int(A) and Cl(A) for sets A containing p and for sets A not containing p.
Excluded point topology is all the subsets of X that exclude p.
where p is in X.

The Attempt at a Solution


So the interior of a set A is the union of all open sets contained in A.
Would the interior for A be A-p , where we exclude p.
and would the interior be A where we include p.
I'm not sure what the smallest closed set would be that contains A.
It seem like it would just be A.

It would be helpful to organize you answers better.

If p∈A,
Determine Int(A).

Determine Cl(A).​

If p∉A,
Determine Int(A).

Determine Cl(A).​
 
  • #3
for the closure of those sets, should I try a proof by contradiction.
for the second one assume that p is not in the closure.Since p is not in A it is in the complement so it is in a closed set.
 
  • #4
Have you proven any properties for closure and interior? For example, you can show that interior of A is the largest open set contained in A and closure of A is the smallest closed set which contains A. If you can prove this, then you're almost done.
 
  • #5
cragar said:
for the closure of those sets, should I try a proof by contradiction.
for the second one assume that p is not in the closure.Since p is not in A it is in the complement so it is in a closed set.
Start with some basics.

If [itex]p\in\text{A}\,,[/itex] then is set A open or is A closed?

If [itex]p\notin\text{A}\,,[/itex] then is set A open or is A closed?
 

1. What is the excluded point topology?

The excluded point topology is a type of topology in which the only open sets are the empty set and the sets that do not contain a specific point, referred to as the "excluded point".

2. How is the interior of a set A defined in the excluded point topology?

In the excluded point topology, the interior of a set A is defined as the set itself, with the excluded point removed. This means that the interior of A is equivalent to the set A without the excluded point.

3. What is the closure of a set A in the excluded point topology?

The closure of a set A in the excluded point topology is the set A itself, along with the excluded point. This means that the closure of A is equivalent to the set A with the excluded point included.

4. How does the excluded point topology differ from other topologies?

The excluded point topology differs from other topologies in that it only has two types of open sets, while other topologies may have more types of open sets. Additionally, the definition of interior and closure may differ in the excluded point topology compared to other topologies.

5. Is the excluded point topology commonly used in real-world applications?

No, the excluded point topology is not commonly used in real-world applications. It is mainly used as an example in topology courses to illustrate different types of topologies and their properties.

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