I have only a superficial acquaintance with topology, but I need to make some use of it in connection with an investigation in another field. I am perplexed by the following line of reasoning, which seems to lead to an erroneous conclusion, but I cannot see the flaw, and would appreciate emlightenment.
Suppose that I have a space X with two sets, P and M, which are disjoint but not separate. That is, they have no points in common but can contain limit points of the other. Let A be a subset of P, and let it be defined as the collection of all points of P that are limit points of M. Now consider the boundary points of A that are members of P and call this set of points B.
1. If xP, and if any neighborhood U
X that contains x also contains some m
M then x
A by definition.
2. Conversely, for any aA, any neighborhood U
X of a contains some m
M because every point of A is a limit point of M.
3. Assume xB, the boundary of A in P. Then any neighborhood S of x contains some other point a
A by definition of boundary.
4. Choose some neighborhood U of a such that US.
5. Then (mM)
U
S. But S is an arbitrary neighborhood of x. Hence x is a limit point of M.
6. Then by (1), xB is a point of A and hence A contains its boundary in P and is closed in P.
(Have I made an error to this point?)
7. Since B is the boundary of A in P, the nighborhood S of xB also contains some y
exterior(A)
P.
8. But x is a limit point of M from (5) and hence every neighborhood S contains some mM and some y
exterior(A) \
P.
9. Hence some y is a limit point of M and therefore a member of A by definition.
Clearly the conclusion that points of the exterior of A are contained in A is absurd. I think that step 5 is legitimate since S converges to x and always contains a point of M. I am less sure about steps 8-9 since as S converges to x, a different y is in S at each step. Still, in the limit there is always some yS and some m
S. Where have I gone wrong?
Later:
I have had a further thought on this. Suppose that X is T2. Then although any S containing x will contain some y and some m, we are guaranteed that we can find some neighborhood containing that y that does not contain x or that m, and there is no guarantee that that neighborhood will contain any other m. Hence if X is T2, 8-9 fails. I believe 5 survives however. Anyone have any thoughts in support or contradiction of this?