Boundary of the union of two sets

[painfive]

is it true that:

$$\partial(A\cup B) = (\partial(A)\cap \mbox{int}(X-B))\cup (\partial(B)\cap \mbox{int}(X-A))$$

? (where $\partial(A)$ is the boundary of A, int(A) is the interior, and A and B are two subsets of the topological space X)

I can prove that:

$$\partial(A\cup B) \subset (\partial(A)\cap (X-\mbox{int}(B)))\cup (\partial(B)\cap (X-\mbox{int}(A)))$$

But I have an example where the equality doesn't hold. (I can show all this if anyone wants). But in the example, the first equalitiy does hold, and it seems like it would always hold, but I can't prove it.

Edit: Actually, now I think I have an example where the first equality doesn't hold. Now I have no idea what $\partial(A\cup B)$ is. (It also isn't $(\partial(A)\cap (X-B))\cup (\partial(B)\cap (X-A))$, because I have a counterexample of that too)

 

[painfive]

I might as well show the counterexample I found. For these two sets, none of the above equalities hold (here X is $\Re^2$):

$$A= {0\le x < 1,0\le y \le 1}$$
$$B= {1\le x \le 2,0\le y \le 1}$$

Here, the first equality should actually be a "contains" and the second two should be "is a subset of".

 

[tacman]

It is not necessarily true that \partial(A\cup B) = (\partial(A)\cap \mbox{int}(X-B))\cup (\partial(B)\cap \mbox{int}(X-A)). This equality only holds if both A and B are open sets or both are closed sets.

To see why, let's consider the definitions of the interior and boundary of a set. The interior of a set A, denoted by int(A), is the largest open set contained in A. The boundary of a set A, denoted by \partial(A), is the set of points that are neither in the interior nor the exterior of A.

Now, if A and B are both open sets, then (\partial(A)\cap \mbox{int}(X-B))\cup (\partial(B)\cap \mbox{int}(X-A)) will contain all the points that are in the boundary of A and B, since both A and B are open and therefore their boundaries are contained in their respective interiors.

Similarly, if A and B are both closed sets, then (\partial(A)\cap \mbox{int}(X-B))\cup (\partial(B)\cap \mbox{int}(X-A)) will contain all the points that are in the boundary of A and B, since both A and B are closed and therefore their boundaries are contained in their respective exteriors.

However, if A is open and B is closed (or vice versa), then this equality will not hold. In this case, the boundary of A\cup B will contain points that are in the interior of A and the exterior of B (since A is open and B is closed), but these points will not be in (\partial(A)\cap \mbox{int}(X-B))\cup (\partial(B)\cap \mbox{int}(X-A)). Similarly, points in the interior of B and the exterior of A will not be in this set either.

So, in conclusion, the equality \partial(A\cup B) = (\partial(A)\cap \mbox{int}(X-B))\cup (\partial(B)\cap \mbox{int}(X-A)) only holds when both A and B are either open or closed sets. In other cases, the boundary of the union of A and B will contain additional points that are not in this set.