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 x P, and if any neighborhood U X that contains x also contains some m M then x A by definition.

2. Conversely, for any a A, any neighborhood U X of a contains some m M because every point of A is a limit point of M.

3. Assume x B, 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 U S.

5. Then (m M) U S. But S is an arbitrary neighborhood of x. Hence x is a limit point of M.

6. Then by (1), x B 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 x B 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 m M 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 y S 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?