Point Set Topology: Why A={1/n:n is Counting Number} is Not a Closed Set?

[kimkibun]

Why is it that the set A={1/n:n is counting number} is not a closed set?

We see that no matter how small our ε is, ε-neighborhood will always contain a point not in A (one reason is that Q* is dense in ℝ), thus, all the elements in A is boundary point, and we know that by definition, if bd(A)≤A, then A is closed (this is what Steven R. Lay used in his book). (≤-subset). A good friend of mine told me that A does not contain cluster point and that made A not a closed set, he said (and i know) that closed set always contain cluster points. is this some sort of contradiction?

 

[micromass]

Sure, all points in A are boundary points. This means that A\subseteq bd(A). What you want is the reverse inclusion! So you have to show that all boundary points are exactly in A. This is not true here, there is a boundary point of A that is not in A.

 

[Bacle2]

Why is it that the set A={1/n:n is counting number} is not a closed set?

We see that no matter how small our ε is, ε-neighborhood will always contain a point not in A (one reason is that Q* is dense in ℝ), thus, all the elements in A is boundary point, and we know that by definition, if bd(A)≤A, then A is closed (this is what Steven R. Lay used in his book). (≤-subset). A good friend of mine told me that A does not contain cluster point and that made A not a closed set, he said (and i know) that closed set always contain cluster points. is this some sort of contradiction?

Re the cluster points, it is true that a closed set contains all its cluster points. Maybe your friend was referring to your set A: does it contain all its cluster points?

 

[HallsofIvy]

As micromass said, the fact that all points in A are boundary points is irrelevant. In order to be closed, all boundary points must be in A. Since this is a sequence of points converging to 0, 0 is as boundary point but is not in A. That is what your friend was saying.

 

[tait]

The set 1/n (n = 1,2,...) doesn't contain any limit points (can you see why?), but it certainly has a limit point (can you see what the limit point is?) and so from the definition we see that this set is not closed.