Metric Space with an Epsilon-Net and Boundedness

[qinglong.1397]

Homework Statement

Prove that if a metric space (X, d) has an \epsilon-net for some positive number \epsilon, then (X, d) is bounded.

Homework Equations

The Attempt at a Solution

I think that (X, d) might be not bounded. For example, let X be a subspace of real line with usual topology. X consists of two intervals. One is, say (-1, 1), and the other is at the infinity with diameter less than 3. Then (X, d) has an 3-net but not bounded, since it does not have a upper bound.

Is there something wrong with my discussion? Thanks a lot!

 

[ystael]

You are correct. For instance, every Euclidean space \mathbb{R}^d has an \epsilon-net for every \epsilon > 0. Some additional condition is needed, such as the requirement that the \epsilon-net be finite.

 

[qinglong.1397]

So, Fred H. Croom made a mistake? Oh, i think it's probability is very low! There must be some mistake in my discussion.

 

[micromass]

Well, all depends of how \epsilon-net is defined.

Does Croom require that \epsilon-nets are finite (like: there are finitely many balls of radius \epsilon that cover)??

If he doesn't require that, then he indeed made a mistake. If he does, then he is 100% correct.


Also, an interval "at the infinity" does not exist. By definition, an interval of R is a set X such that
\forall x,y\in X:~x\leq z\leq y~\Rightarrow~z\in X
Since X is required to only contain real numbers, infinity can never have infinity in it. Thus an interval at infinity does not exist...

 

[qinglong.1397]

Well, all depends of how \epsilon-net is defined.

Does Croom require that \epsilon-nets are finite (like: there are finitely many balls of radius \epsilon that cover)??

If he doesn't require that, then he indeed made a mistake. If he does, then he is 100% correct.Also, an interval "at the infinity" does not exist. By definition, an interval of R is a set X such that
\forall x,y\in X:~x\leq z\leq y~\Rightarrow~z\in X
Since X is required to only contain real numbers, infinity can never have infinity in it. Thus an interval at infinity does not exist...

Since you define this interval in real line: \forall x,y\in X:~x\leq z\leq y~\Rightarrow~z\in X, then if z=infinity, this interval contains some elements which are all infinitely large. Doesn't that mean this interval is in the infinity? Do you mean infinitely large number is not real? It sounds strange.

 

[micromass]

Well, then it all depends whether infinity is a real number. The convention that mathematicions make is that infinity is not real. Thus we can't take an interval on infinity, simply because infinity doesn't exist (when talking about the reals).

What you can do however, is extend the real numbers with infinity. This gives us \mathbb{R}\cup\{+\infty,-\infty\}:=\overline{\mathbb{R}}. These are called the extended real numbers. These numbers are a good setting for limits and integrals. And it can even be given a metric space structure by

d(x,y)=|atan(x)-atan(y)|

where atan is extended to infinity to yield the value \pm \pi/2. But note that this metric space is bounded!

Sorry for my digression. The point that I wanted to make is that infinity is not a real number...