# Boundedness in a topological space?

1. I was wondering if there is a concept of boundedness for subsets of a topological space?
2. If yes to 1, is it this one from Wiki

Elements of a Bornology B on a set X are called bounded sets and the pair (X, B) is called a bornological set.

For any topological space X, the set of subsets of X with compact closure is a Bornology.

3. If yes to 2, does it coincide with boundedness in a metric space and in a topological vector space? How is it related to total boundedness in a uniform space?

Thanks and regards!

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If you have found the answer helpful and correct, please accept it else let people know what more do you want from the answer. :-) – Jayesh Badwaik Aug 4 '12 at 7:12

In general, there is no notion of boundedness on a topological space.

Exercise: given a metric space $(X,d)$, show that $D(x,y):=\mbox{min}(1,d(x,y))$ defines a second metric on $X$ which is equivalent to the first one; that is, a subset $U$ of $X$ is open with respect to the first metric if and only if it is open with respect to the second one.

Note that with the second metric, every set is bounded, but the topologies are the same. What this shows is that, in metric spaces, where the notion of boundedness is well-defined, one can show that it is in fact independent of the topology. Boundedness is a property which arises from the metric.

The concept of bornology (although I am not familiar with it) allows you to study boundedness by adding extra structure to your topological space, and what you get is a bornological space.

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+1 Thanks! Is the case for topological vector spaces similar to that for metric spaces? – Tim Feb 26 '12 at 16:07
I'm not sure what your question is, but I'll add this. To define boundedness on topological vector spaces, you're using the extra structure: either the semi-norms used to define the topology, or in general the scalar product. The point I was making is that a bornology is a way to abstract the notion of boundedness which is available in some contexts (metric spaces, top. vector spaces). But you can't study boundedness without adding extra structure. – M Turgeon Feb 26 '12 at 16:16
@MTurgeon : with the second metric, every set is bounded Could you explain why? For example, let $X=\mathbb R$, $U = \mathbb R$. Then $U$ isn't bounded, since it isn't included in any $(-r,r)$ open ball. Didnt you mean $D(x,y):=\mathrm{min}(1,d(x,y))$? Or, have I misunderstoot something? – mma Jun 28 '15 at 9:44
@mma You're absolutely right, I can't believe this typo has been there all this time... – M Turgeon Jun 28 '15 at 14:17
@mma In a discrete metric space, all sets are closed and bounded, and a set is compact if and only if it is finite. Therefore, all infinite sets are bounded with non-compact closure. – M Turgeon Jun 29 '15 at 16:11

To see the relation between boundedness, total boundedness, and kinds of topological spaces, I suggest you consider characterizations of compactness.

An example that helps me here is the one-point compactification of the real line. If we take away the metric, the point at infinity is just like all the other points. So there's no boundedness or unboundedness.

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The bornology defined in the Wiki article you referenced is normally called a hereditary ideal in set theory, although a hereditary ideal is not required to cover the space. However, as ideals are used to define notions of "smallness", and usually points are "small", all of the widely studied ones I'm aware of do cover the space. Examples are the Lebesgue measure zero subsets of the line, and the meager sets in a topological space.

An example of a hereditary ideal that doesn't cover the space is the set of $\mu$ measure zero sets where $\mu$ gives at least one point positive measure.

I've never heard of a hereditary ideal being used to define a notion of boundedness.

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Why hereditary? From a set-theoretic point of view it’s just an ideal in $\langle\wp(X),\subseteq\rangle$ that covers $X$. – Brian M. Scott Feb 27 '12 at 4:44