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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.

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2 Answers 2

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.

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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.

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