If P is the set of all isolated points of S, then P is a closed set.Let S and T be a subsets of R. Find a counterexample. 

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

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Consider the set S of points:

`{(1)/(2^n)} _(n in N)`

All of these points are isolated.  To see this, for any point in S, we need to find a neighborhood that contains no other points of S. Consider the interval:

`(1/(2^n)-1/(2^(n+2)), " " 1/(2^n)+1/(2^(n+2)))`

That interval contains the point `1/2^n`

But it doesn't contain any other point of S:

`1/2^n+1/2^(n+2)=1/2^n+(1/4) (1/2^n) =(5/4) (1/2^n) < (2) (1/2^n)=1/2^(n-1)`

`1/2^n - 1/(2^(n+2)) = 1/2^n - (1/4) (1/2^n) = (3/4) (1/2^n) > (1/2) (1/2^n) = 1/(2^(n+1))`

That is, the next closest points to `1/2^n` are excluded from that interval.  So there is an open neighborhood (interval) containing `1/2^n` and not containing any other point of S.  So every point in S is isolated.  

BUT S is not closed.

A closed set must contain all of its limit points.  Clearly 0 is a limit point of points of S, so if S were closed, it would have to contain 0.  Thus S is a counterexample to the statement

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