# Thread: Topology Boundry Question

1. ## Topology Boundry Question

For each $\displaystyle n\in Z_+$, let $\displaystyle B_n = {n, n+1, n+2, ....}$ and consider the collection B = $\displaystyle {B_n | n\in Z_+}$

(a) Show that B is a basis for a topology for $\displaystyle Z^+$

(b) Show thte the topology on X generated by B is not Hausdorff

(c) Show that the sequence (2,4,6,8,...) converges to every point in $\displaystyle Z_+$ with the topology generated by B

2. Originally Posted by flaming
For each $\displaystyle n\in Z_+$, let $\displaystyle B_n = {n, n+1, n+2, ....}$ and consider the collection B = $\displaystyle {B_n | n\in Z_+}$

(a) Show that B is a basis for a topology for $\displaystyle Z^+$

(b) Show thte the topology on X generated by B is not Hausdorff

(c) Show that the sequence (2,4,6,8,...) converges to every point in $\displaystyle Z_+$ with the topology generated by B
(a) The followings are the basis elements of B.

B1={1,2,3,.................}
B2={2,3,4,...............}
B3={3,4,5,............}
........
Bn={n, n+1, n+2,,,,}
...........

To check B is a basis for Z+, you first need to check the union of the above elements are indeed Z+.
Second, you need to verify that for each x in the interesection of the basis elements, there exists a basis element containing x and contained in the intersection.

For instance, consider the intersection of B1 and B2. For each x in the section, you can find a basis element that containing x.

(b) Pick two numbers, for instance, 3 and 5. Check open sets containing 3 & 5 respectively. Those open sets cannot be chosen disjointly.

(c) Every open set containing a number in Z+ contains all but finite numbers of the above sequence in your topology. Thus, it converges everywhere in Z+.