If S is any subset of E 0, then S ¯ denotes the smallest hereditary saturated subset of E 0 which contains S ; S ¯ is called the hereditary saturated closure of S. (Such exists by the previous observation).
Let H be the smallest hereditary saturated subset of (E^0) that contains all the cycles of E. Order the vertices and the cycles by the preorder defined in Remark 5.9 (1).
On the other hand, by our construction of the graph (E"), it inherits all the maximal cycles of E, which are all non-exclusive, and ((E")^0) is equal to the smallest hereditary saturated subset (with respect to (E")) containing all the cycles.
Indeed, the set of all descendants of (w_1) is a hereditary subset of V.
Remark that each component (S_i) intersects just one minimal hereditary subset (W_i).
Let W be a nonempty hereditary subset of V. Let (W^bot subset V) consist of those vertices which do not have descendants in W. Clearly, (W^bot ) is a hereditary subset of V.
Indeed, if the Leavitt path algebra (L(Gamma )) is prime then the frame consists of one minimal hereditary subset (W_1).
Any subset of U is a hereditary subset of C N 0. We note also that, since saturation applies only to regular vertices, any subset of U is saturated as well.
Let W be a minimal nonempty hereditary subset of V. Then the ideal I(W) is generated (as an ideal) by all idempotents (e_w, w in W).
Let W be a minimal nonempty hereditary subset of V. Then for any two vertices (w_1,W_2in W) the vertex (w_2) is a descendant of (w_1).
Indeed, the only minimal hereditary subset of V is ({w}.) There are infinitely many special paths with all vertices lying in (V{setminus }{w}={v}.) Let (Gamma = ) Open image in new window The graph (Gamma = V,E)) does not have proper hereditary subsets.
