**Exercise**
Let *A* be a subset of a topological space *X*.
How many different subsets can I make starting from *A* and
repeatedly using the boundary and union operations?

Answer: 13.

Let us write d for boundary and + for union, so that
the closure of *A* is written *A+dA*.
Now the thirteen sets are
*A*, *A+dA*, *A+ddA*, *A+d(A+dA)*,
*A+d(A+ddA)*, *A+dd(A+ddA)*, *A+d(A+d(A+ddA))*,
*dA*, *ddA*, *d(A+dA)*, *d(A+ddA)*,
*dd(A+ddA)*, *d(A+d(A+ddA))*.
(A.E. Brouwer, 1969)

**Proof**

The example shows that one can get thirteen different subsets.
It also suggests what we have to prove.
Let us shorten notation a bit and write *B = A+ddA*
so that the six boundaries become *dA*, *ddA*, *d(A+dA)*,
*dB*, *ddB*, *d(A+dB)*.

(i) *ddC = dC* *for any closed set* *C*
(in particular for *C = dA* and *C = A+dA*).

Indeed, if *C* is closed then *dC* is contained
in *C*, so the interior of *dC* is contained in
the interior of *C* which is disjoint from *dC*,
so *dC* has empty interior.

(ii) *If* *F* *is contained in* *dA*,
*then* *d(A+F)* *contains* *d(A+dA)*
*and is contained in* *dA*.

Indeed, *d(A+F)* is contained in the closure of *A*
and is disjoint from the interior of *A*, so lies in *dA*.
On the other hand, *d(A+dA)* is contained in *d(A+F)*
because it is contained in the closure of *A+F*
(which is the closure of *A*) and is disjoint from the interior
of *A+F* (which is contained in the interior of *A+dA*).

(iii) *If C is closed, and D\C = E\C, then dD\C = dE\C*.

Indeed, Int*(D)\C = *Int*(D\C)*.

(iv) *ddA + dB = dA* and *ddB* is contained in *ddA*
and Int*(dA) = *Int*(dB)*.

Indeed, Since *B\ddA = A\ddA* we find from (iii) that
*dB\ddA = dA\ddA*. Applying (iii) once more, we see
*ddB\ddA = ddA\ddA*. Finally, *dB* is contained in *dA*
so Int*(dA)* is contained in Int*(dB)*, while on the other
hand Int*(dA) = dA\ddA = dB\ddA* is contained in
*dB\ddB = *Int*(dB)*.

(v) *d(A+d(A+dA)) = dA*.

Indeed, by (ii) the LHS is contained in the RHS. By (iii)
it suffices to show that the LHS contains *d(A+dA)*,
and this follows from (ii).

(vi) *d(A+ddB) = dA*.

Indeed, by (ii) the LHS is contained in the RHS. By (iii)
it suffices to show that the LHS contains *ddB*.
But the interior of *A+ddB* is contained in the interior
of *A+ddA = B* which is disjoint from *dB* and hence
from *ddB*.

(vii) *ddB + d(A+dB) = ddA*.

Indeed, *d(A+dB)* is contained in *ddA* since
Int*(dA)* is contained in Int*(A+dB)* by (iv).
On the other hand, a point of *ddA* not in the LHS
lies in *dA* but not in Int*(dA) = *Int*(dB)*,
so not in *dB* but in Int*(A+dB)*, hence in Int*(A)*,
contradiction.

(viii) *dd(A+dB) = d(A+dB)*.

Indeed, a point in the interior of *d(A+dB)* lies in the
interior of *ddA* which is empty.

(ix) *d(A+d(A+dB)) = dB*.

Indeed, using (iii) twice we see that *d(A+dB)\dB = dA\dB*
and *(A+d(A+dB))\dB = (A+dA)\dB* and
*d(A+d(A+dB))\dB = d(A+dA)\dB* which is empty by (ii).
On the other hand, a point of *dB* that is not in the LHS
lies in the interior of *A+d(A+dB)*, hence in the interior
of *A+ddA = B*, contradiction.

This settles everything.