Give an exampleSolutiona Example of a closed subset of R and

Give an example:

Solution

a) Example of a closed subset of R and a point a in S which is not a cluster point of S.

A cluster point a, of a set A, is a point, within the set A or outside, such that every deleted neighbourhood of a must contain atleast one element of the set A other than a.

So let us consider the set N of all natural numbers. Then N is a subset of R. In order to search for the cluster points of N, we see that N is discrete in nature, i.e. for each n in N, we have a neighbourhood (-p,p) for some arbitrarily small real no. p, such that the neighbourhood has no points of N other than p.

So clearly n is not a cluster point of N and this goes for each n in N. So we can conclude that N has no cluster points.

Now to claim that N is closed, we must show that each cluster point (if any exist) belong to N. Since there doesn\'t exist any cluster point of N, the property of being closed subset is vacuously satisfied. Thus we conclude that N is a closed subset of R which has no cluster point. So to be precise to the question, take S=N and a=1. Clearly then S is closed subset of R and a is not a cluster point of S.

These kind of sets are called discrete sets.

ii) Example of subset of R, the set of all real numbers which is neither open nor closed but each point of S is a cluster point of S.

We consider the set Q of all rational numbers. Q is a subset of R.

To show that Q is not open, we have to show that no point of Q is an interior point of Q, i.e. there is no rational number such that which has atleast one neighbourhood of points contained in Q. Now this happens because becuase for every pair of rationals, there exist infinity many irrationals. So for each q in Q and each positive real p(arbitrarily small), the neighbourhood (q-p,q+p) must contain an irrational number. So no point of Q is an interior point. So Q is not open in R.

Now the density property of Q shows that between every pair of rational numbers, there are infinitely many rationals. Thus we can conclude that each rational number is a cluster point of Q.

Now to show that Q is not closed in R, we have to show that Q do not contains all its cluster points, i.e. there are points outside Q, such that each neighbourhood of that point intersect Q.

Take an irrational number r from R and a positive real number a (arbitrarily small). Then (r-a,r+a) is a neighbourhood of r. Now for this r, we have a sequence of rational numbers {q_n} which converges to r. So the neighbourhood (r-a,r+a) must contain an infinite number of rational numbers in it, which is equivalent to the fact that r is a cluster point of Q. But r is not in Q.

So it is clear that Q has cluster points outside Q and hence it is not closed in R.

The above claim is true for every irrational number. So we can conclude that every irrational number is a cluster point of Q.

This shows that Q is neither a closed subset of R nor open still every point of R is a cluster point of Q.