- #1
Adriadne
- 23
- 0
I posted this on another forum, but had no response. Maybe because it's too stupid the bother with? Anyway...
Say I have a set X and a topology T on X so that T = {X {} A} i.e A is an open subset of T. Then the complement of A is Ac = X - A, which is closed.
Now the interior of A, int(A) is the largest open set (or the union of all open sets) contained in A which is A, and the closure of A, cl(A) is the smallest closed set in {X {} Ac} containing A which is X. So if the boundary of A
bd(A) = cl(A) - int(A), we have that bd(A) = X - A = Ac.
Similarly, the closure of Ac is the smallest closed set containing Ac, which is Ac = X - A. So, using the alternative definition for the boundary of A,
bd(A) = cl(A) intersect cl(Ac) = X intersect (X - A) which is X - A = Ac.
This argument applies also for the trivial and discrete topologies, where the boundaries are respectivley X and {}. I've also tried it out on a number of arbitrary topologies of my own devising, and the answer is always the same, the boundary of A is the complement of A. Surely it's not right, though?
Say I have a set X and a topology T on X so that T = {X {} A} i.e A is an open subset of T. Then the complement of A is Ac = X - A, which is closed.
Now the interior of A, int(A) is the largest open set (or the union of all open sets) contained in A which is A, and the closure of A, cl(A) is the smallest closed set in {X {} Ac} containing A which is X. So if the boundary of A
bd(A) = cl(A) - int(A), we have that bd(A) = X - A = Ac.
Similarly, the closure of Ac is the smallest closed set containing Ac, which is Ac = X - A. So, using the alternative definition for the boundary of A,
bd(A) = cl(A) intersect cl(Ac) = X intersect (X - A) which is X - A = Ac.
This argument applies also for the trivial and discrete topologies, where the boundaries are respectivley X and {}. I've also tried it out on a number of arbitrary topologies of my own devising, and the answer is always the same, the boundary of A is the complement of A. Surely it's not right, though?