Question

Prove that the set of natural numbers n is closed set

Answer 1

Step-by-step explanation:

That set is the union of an infinite number of open intervals, so it is open. That makes the natural numbers closed. ... In the topological sense, yes, the integers are a closed subset of the real numbers. In topological terms, it means that, for any real number that is not an integer, there is an “open set” around it.

Answer 2

Step-by-step explanation:

To prove that it's closed, we will prove that it's complement is open. Consider the sets Si=B(i+1/2,1/2)Si=B(i+1/2,1/2), where B(x,r)B(x,r) is the open ball of radius rrcentered at xx. Each SiSi is open.

(−∞,0)∪(∪i∈NSi)=R−N(−∞,0)∪(∪i∈NSi)=R−N

so R−NR−N is the countable union of open sets and so it's open. Therefore complement of R−NR−N, which is NN, is closed.

To prove that it's closed we will prove that it's complement is open. Let x∈R−Nx∈R−N. We wish to show that there is an open ball centered at xx that contains no integers. Let [x][x] denote the distance from xx to the nearest integer. This is either the fractional part of xx or one minus the fractional part of xx, whichever is smaller. Then the ball B(x,[x]/2)B(x,[x]/2) doesn't contain any integers. Thus