Counterexample?

**symbol0** · Jun22-09, 11:03 PM

Hi, I read the following theorem in a book:
If {

} is a sequence of nowhere dense sets in a complete metric space X, then there exists a point in X which is not in any of the

's.

But what if I say X={1,1/2,1/3, ...} $LaTeX Code: \\cup$ {0} with the regular metric d(x,y)=|x-y|, and

={1/n,0}.
Why wouldn't this be a counterexample?
As I see it, X is a complete metric space, each

is nowhere dense, and their union equals X.

**gel** · Jun23-09, 03:03 AM

A_n is dense at 1/n

**symbol0** · Jun23-09, 10:25 PM

I don't understand what you mean by "An is dense at 1/n ".
The definition I have is: A subset S of a metric space is said to be nowhere dense if its closure has empty interior.
The closure of each set in the sequence I propose, {1/n,0}, is the set itself. These sets have empty interiors because 1/n and 0 are not interior points of {1/n,0}.

**Hurkyl** · Jun23-09, 10:50 PM

Originally Posted by symbol0

1/n and 0 are not interior points of {1/n,0}.

Why not?

**symbol0** · Jun23-09, 10:55 PM

Because no open ball centered at 1/n or 0 belongs to {1/n,0}

**Hurkyl** · Jun23-09, 11:13 PM

Originally Posted by symbol0

Because no open ball centered at 1/n or 0 belongs to {1/n,0}

Are you sure about that? What about the one with radius 1/(2n)?

**symbol0** · Jun24-09, 12:11 AM

A ball of radius 1/2n would contain points that are not in the set. For example, take the set {1/3,0}. A ball of radius 1/6 centered at 1/3 would contain the point 1/4 which is not in {1/3,0}. So the ball does not belong to the set. The same goes for all sets {1/n,0}.

**Hurkyl** · Jun24-09, 01:00 AM

Originally Posted by symbol0

A ball of radius 1/6 centered at 1/3 would contain the point 1/4 which is not in {1/3,0}.

Hrm. Yep, I made an error. But it is obvious how to fix my statement, right?

**symbol0** · Jun24-09, 11:56 PM

I think I see now why the points 1/n and 0 are interior points of each set.
The problem was that all this time I've been under the assumption that a set with only one point could not be an open ball (this happens when the metric space is $LaTeX Code: \\mathbb{R}$ , but it does not happen in a metric space like the one I am proposing). So if I center balls at 1/n and 0 with a very small radius, then the only point of the ball that is inside the set will be the center of the each ball. So the ball is inside the set (it does not matter that the ball has only one point).
Thus 1/n and 0 are interior points.
So the sequence I proposed is NOT a sequence of nowhere dense sets.

Am I right?

**Hurkyl** · Jun25-09, 10:26 AM

Originally Posted by symbol0

So if I center balls at 1/n and 0 with a very small radius, then the only point of the ball that is inside the set will be the center of the each ball.

Correct for 1/n. Incorrect for 0. (Every ball, no matter how small of a radius, does contain another point) While 1/n is, 0 is not an interior point of any of your sets.

**symbol0** · Jun25-09, 09:53 PM

That's right Hurkyl.
Thanks for your help.