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

 

[mighty2000]

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.