Let $(X,\leq)$ be a well-ordered set, Let $\mathscr T$ be the order topology on $X$. Then $(X,\mathscr T)$ is normal.

Proof. We know that $(X,\mathscr T)$ is Hausdorff. Then, It is $T_1$. Every interval in $X$ is of the form $(a,b]$ is open in $X$.

Since $(X,\leq)$ is well-ordered set, $X$ has a least member $a_0$. Suppose $C$ and $D$ are disjoint closed subsets of $X$.

Our aim is to find two disjoint open subsets of $X$, $U$ and $V$.

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How does the underlined statement be true?

My effort:-

Since $D$ is closed, $X\setminus D$ is open. We also know that $C\cap D=\emptyset$. So, $C\subset X\setminus D$. I know that, $\forall a\in C, a\ge a_0.$ Also there is an open set $U_a$ containing $a$ which is a subset of $X\setminus D.$ I don't know anything other than this. How do I prove there is an interval $(x_a,a]$ such that $(x_a,a]\cap D=\emptyset?$


1 Answer 1


We know there is an open set $U_a$ containing $a$ that misses $D$.

Now use that we have the order topology (!): a base for that topology consists of all open intervals $(x,y)$ with $x,y \in X, x < y$ and also, if $\min(X)$ exists, all sets of the form $[\min(X), x)$ and if $\max(X)$ exists all sets of the form $(x,\max(X)]$ as well (otherwise $\min(X)$ or $\max(X)$ would not be in any basic open set, if we only allowed the open intervals).

So if $a$ happens to be $\max(X)$ we are done right away, as we have a basic element $(c,a]$ inside $U_a$.

If not, there must be some $(c,d)$ containing $a$, that sits inside $U_a$ by the definition of a base.

Now we use the well-order: if $(a,d)$ is empty, we have no points between $a$ and $d$ and then $(c,d)=(c,a]$ and we are done. If it is non-empty: it has a minimum that we call $a^+$ (the smallest element larger than $a$) and then $a \in (c,a^+] \subseteq (c,d) \subseteq U_a$ and we are done too. (we pick $x_a=c$ in all these cases)


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