Question

(4) Let No = NU{o} be the union of the set of natural numbers with a single point, called oo. Give No the order which is the natural order on points of N, and extend this order to the point o by: VnEN, n<o. Give Noo the order topology. What are the limit points of N..?

Answer 1

We have the set,
N = NU
along with order topology given by the natural order. That is for
m,n EN
we have the natural ordering of integers and we have set
no
for all
neN
. Then $\mathbb{N}_\infty$ is given the order topology with respect to this order.

We are to find out the limit points of this topological space.

We recall the definition first. A point $x\in X$ of a topological space is called a limit point of if for every open set
IEUX
containing we have a point
Y EU {
, i.e, a point other than .

Let us first show that for a natural number
neN
where $n \ne 1$ , it is not a limit point of $\mathbb{N}_\infty$ . For this consider the set,
Un = {mENn-1<m<n +1}
. In the order topology, $U_n$ is clearly open. We have,
Un = {n}
and hence cannot be a limit point.

For $n=1$ , consider the set
V1 = {m € Nm <2} = {1}
. Again in the order topology, $U_1$ is open. Hence, $n=1$ is not a limit point.

We are left with the point $\infty$ . We claim that this is indeed a limit point. In the order topology, a sub-basic open set containing $\infty$ looks like,
Vn = {m eNm >n
for some
neN
. Suppose is some arbitrary open set containing $\infty$ . Then we must have,
100 E no cu
for some $n_0$ . But then clearly,
no +1 €Uoo
. Hence we see that $\infty$ is a limit point.

Hope this helps. Feel free to comment for any further clarifications. Cheers!