Find Boundary of A (-1,1) U {2} Lower Limit Topology

• Fluffman4
In summary, the interior of A in the lower limit topology on R is [0,1) and the boundary is {2}. The closure of A is A itself since A is already closed in this topology.
Fluffman4
Determine the boundary of A.

A= (-1,1) U {2} with the lower limit topology on R

What I know is that the topology defines open sets as those of the form [a,b). In this case, if they want an interval in the form of [a,b) for the interior, then it comes to mind that [0,1) would be the interior since [-1,1) is not contained in A. The closure of the set also baffles me a little, because I'd think that would just be [-1, 2) since R - [-1, 2) = (-infty, -1) U [2, infty) which is an open set in R with the lower limit topology, so I'd figure that [-1, 2) is closed in the lower limit topology.

Is it possible that anybody can help me to figure out the interior of A or at least push me in the right direction?
Thanks.

I would approach this problem by first understanding the definition of the lower limit topology on R. This topology is generated by the basis B = {[a,b) | a,b ∈ R, a < b} which means that any open set in this topology can be written as a union of basis elements.

Next, I would look at the set A and determine which basis elements are contained in it. In this case, A contains the basis element [0,1) but not [-1,1). Therefore, the interior of A would be [0,1).

To determine the boundary of A, I would consider the complement of A in R, which is (-∞,-1] U [1,2) U (2,∞). The boundary of A would then be the set of points where every neighborhood intersects both A and its complement. In this case, the only point that satisfies this condition is 2, so the boundary of A is {2}.

In terms of closure, the closure of A would be the smallest closed set that contains A. Since A is already closed in the lower limit topology, the closure of A would simply be A itself.

I hope this helps to clarify the boundary and closure of A in the lower limit topology on R. Let me know if you have any further questions or need more clarification.

1. What is the definition of "Find Boundary of A (-1,1) U {2} Lower Limit Topology"?

The boundary of a set in a topological space is the set of points that are both in the closure of the set and in the closure of its complement. In this case, the set A is defined as the open interval between -1 and 1, union with the point 2, in the lower limit topology.

2. How is the boundary of A (-1,1) U {2} Lower Limit Topology calculated?

To calculate the boundary of A (-1,1) U {2} Lower Limit Topology, we first take the closure of A, which is the closed interval between -1 and 1, union with the point 2. Then, we take the closure of the complement of A, which is the open interval between -∞ and -1, union with the open interval between 1 and ∞. The boundary is the intersection of these two sets, which is the closed interval between -1 and 1, union with the point 2.

3. What are the points that lie on the boundary of A (-1,1) U {2} Lower Limit Topology?

The points that lie on the boundary of A (-1,1) U {2} Lower Limit Topology are -1, 1, and 2. This is because these points are both in the closure of A and in the closure of the complement of A.

4. Is the boundary of A (-1,1) U {2} Lower Limit Topology a closed set?

Yes, the boundary of A (-1,1) U {2} Lower Limit Topology is a closed set. This is because it contains all of its limit points, and in a lower limit topology, the closure of a set is equal to the set itself plus its limit points.

5. How does the boundary of A (-1,1) U {2} Lower Limit Topology relate to the concept of separation?

In topology, a set is considered to be separated if it does not contain its boundary. In this case, the boundary of A (-1,1) U {2} Lower Limit Topology is non-empty, which means that the set is not separated. This is because the points on the boundary are shared by both the set A and its complement, making it impossible to separate them.

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