Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to see any linear order space (LOTS) is regular? In other words, is it always regular?

Thanks for your help. Any help will be appreciated.

share|cite|improve this question
up vote 3 down vote accepted

Although, as Brian Scott pointed out, regularity is an immediate consequence of stronger separation properties of LOTS, it seems worthwhile to note that, if one only wants regularity, then there are easier proofs. Suppose $x$ is a point and $C$ is a closed set not containing $x$. Then $x$ is in an open interval $(a,b)$ disjoint from $C$. If there are points $p$ and $q$ with $a<p<x<q<b$, then $(p,q)$ is an open neighborhood of $x$, whose closure, being included in $[p,q]$ and therefore in $(a,b)$, is disjoint from $C$. It remains to consider the cases where $x$ is the smallest element of $(a,b)$, or the largest, or both.

The case of "both" is trivial, as then $(a,b)=\{x\}$ is both open and closed. Suppose, therefore, that $x$ is the smallest but not the largest element of $(a,b)$. So we have $q$ with $a<x<q<b$, but there are no points between $a$ and $x$. Then $(a,q)$ is an open neighborhood of $x$. Its closure is included in $[a,q]$, so if we show that the closure doesn't contain $a$, then we'll have the closure included in $(a,b)$ and thus disjoint from $C$ as required. But $(-\infty,x)$ is a neighborhood of $a$ disjoint from $(a,b)$ because there are no points between $a$ and $x$; therefore $a$ is indeed not in the closure of $(a,q)$, and this case of the proof is complete. The remaining case, where $x$ is the largest but not the smallest element of $(a,b)$, is treated symmetrically.

share|cite|improve this answer

I proved a stronger statement here: every LOTS is hereditarily normal. And in his answer Henno Brandsma gives links to a proof that every GO-space is monotonically normal.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.