# Closures on ordered sets

1. Sep 13, 2010

1. The problem statement, all variables and given/known data

OK, this question may sound somewhat trivial, but I'd still like to check my answer. It's from Munkres.

Let X be an ordered set with the order topology. One needs to prove that Cl(<a, b>) is a subset of [a, b] and investigate the conditions necessary for equality.

3. The attempt at a solution

So, if x is any element in <a, b> it is obviously in [a, b], too. If it happens that either a or b (or both) are in Cl(<a, b>), they are in [a, b] too, so Cl(<a, b>) is a subset of [a, b].

Now, if x is any element of [a, b] different from a and b, obviously x is in <a, b>, too, which is a subset of Cl(<a, b>), and hence x is in Cl(<a, b>). Now, the interesting cases are a and b. Is a in Cl(<a, b>)? Well, if so, then every neighborhood of a intersects <a, b>. Let X be a finite ordered set. Is a is the least element of X, then every interval of the form [a, b> intersects <a, b>, and it's obviously the only type of neighborhood that contains a. But if a is not the least element of X, then <a1, a, a2> (where a1 < a and a2 is the first element such that a2 > a) is a neighborhood containing a which doesn't intersect <a, b>. So, a can't be in Cl(<a, b>) if X is finite, unless a is the least element of X. An analogous proof holds for b.

Now, if my reasoning above is correct, it's natural to ask the question: are a and/or b in Cl(<a, b>) if X is infinite? Clearly, motivated by the arguments above, one can reason that between every element which is the first one greater than a and a itself there should be another element, and so on, so we could chose an open neighborhood of a which intersects <a, b>. So X should be infinite. But does every infinite ordered set satisfy this condition?

I'm a bit confused on this one, thanks for any help.

2. Sep 13, 2010

Oops, it just occured to me that, if X has a smallest element a, {a} is a neighborhood of a which doesn't intersect <a, b> (since the smallest half-interval which contains a is {a}). So, actually, if X is finite, a and b can not belong to Cl(<a, b>). Still, my last question remains open.

3. Sep 13, 2010

### Eynstone

The inclusion can be strict - consider the least uncountable well-ordered set.

4. Sep 13, 2010