Why is r/2 used in the proof for one point set being closed?

[radou]

So, I'm going through a proposition, which states that if (X, d) is a metric space, then any set {x}, where x e X, is a closed subset of X.

First of all, could we do this proof to assume the contrary? Since then obviously for the point x from {x} there doesn't exist any real number r > 0 such that the open ball K(x, r) is contained in {x}?

The proof in the notes I'm going through relies on the fact that we have to prove that the complement of {x}, i.e. X\{x} is open. The proof is very simple too, although I'm not quite sure about one thing. Let x' be an element of X\{x}. Then d(x', x) = r > 0, so the open ball K(x', r/2) is contained in X\{x}, and if we take the union for all x' e X\{x} of all such open balls, we get X\{x}, and hence X\{x} is open.

Now, why is it r/2 ? Wouldn't open balls of type K(x', r) be contained in X\{x} too, since K(x', r) = {x'' in X : d(x''-x') < r}, and this set can't contain x, since d(x', x) = r? Perhaps I'm missing something trivially obvious here?

 

[some_dude]

It's simpler if you just do it directly:

y in X\{x} => d(x, y) > 0

Then use that to show X\{x} is open.

 

[radou]

That's just the proof I demonstrated, isn't it? The question about r/2 and about the proof by contradiction still remains unanswered.

Although, in this "proof by contradiction", we can just show that {x} can't be open, but this doesn't necessary imply that {x} is closed? Since the definition of closed is that its complement is open. And in this proof, we didn't show anything about its complement.

So, I guess the one from the text is the only proof? If what I wrote above is correct, then I only want to know is the proof would work with r instead od r/2.

 

[eok20]

First of all, could we do this proof to assume the contrary? Since then obviously for the point x from {x} there doesn't exist any real number r > 0 such that the open ball K(x, r) is contained in {x}?

That proves that {x} is not open. But a set being not open does NOT imply that the set is closed (e.g. [0,1) as a subset of R is neither open nor closed).

 

[eok20]

The proof in the notes I'm going through relies on the fact that we have to prove that the complement of {x}, i.e. X\{x} is open. The proof is very simple too, although I'm not quite sure about one thing. Let x' be an element of X\{x}. Then d(x', x) = r > 0, so the open ball K(x', r/2) is contained in X\{x}, and if we take the union for all x' e X\{x} of all such open balls, we get X\{x}, and hence X\{x} is open.

Now, why is it r/2 ? Wouldn't open balls of type K(x', r) be contained in X\{x} too, since K(x', r) = {x'' in X : d(x''-x') < r}, and this set can't contain x, since d(x', x) = r? Perhaps I'm missing something trivially obvious here?

You are correct that you can use r as well. Sometimes people use lower things to be safe.

 

[radou]

That proves that {x} is not open. But a set being not open does NOT imply that the set is closed (e.g. [0,1) as a subset of R is neither open nor closed).

Yes, that's what I just realized, thanks.

You are correct that you can use r as well. Sometimes people use lower things to be safe.

OK, thanks! Although, if we're being completely rigorous here, I don't see any additional "safety" in it. :)

 

[Hurkyl]

OK, thanks! Although, if we're being completely rigorous here, I don't see any additional "safety" in it. :)

It's "safe" in the sense that you don't have to think about it at all. Using r, you have to think about the (literal ) edge case before you can be satisfied with the proof. Using r/2 you don't have to think about it at all.

Okay, in this case there isn't much to think about, but after you do it for a while, it becomes habit to simply make things smaller to render fine detail irrelevant.