I just finished reading through the section about normal spaces (resp. $T_4$-spaces) from Willard General Topology text, and came across the following problem:
Show that the Scattered Line (also called the Michael Line) is $T_4$.
I'm not sure how to prove this, let alone get started on the problem. Any help would be much appreciated.
Note: The Scattered Line (also called the Michael Line) is given as follows: Let $\mathbb{R}$ be the real number line. Let $\mathbb{P}$ be the set of all irrational numbers. Let $\mathbb{Q}=\mathbb{R}-\mathbb{P}$, the set of all rational numbers. Let $\tau$ be the usual topology of the real line $\mathbb{R}$. The following is a base that defines a topology on $\mathbb{R}$.
$$\mathcal{B}=\tau \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}$$
The real line with the topology generated by $\mathcal{B}$ is called the Scattered Line (also called the Michael Line).
Best Answer
This can be viewed (as Brian M. Scott also does) from a more general point of view:
A $T_1$ space is called monotonically normal iff there is a so-called $\mu$-operator that assigns an open subset $\mu(x,O)$ for every pair $x,O$ where $x \in O$ and $O$ is open in $X$ and so that
This looks rather innocent but is in fact quite powerful: it's a hereditary property: if $X$ is monotonically normal and $A \subseteq X$ then so is $A$ in its subspace topology: if $O \cap A$ is an open subset of $A$ (so $O$ is open in $X$) we define $\mu_A(x, O \cap A) = \mu(x,O) \cap A$, when $\mu$ is the operator on $X$.
Fact
Proof: as being monotonically normal is hereditary, it suffices to show $X$ is normal (the $T_1$ is part of the definition). So let $A$ and $B$ be disjoint closed sets of $X$.
Define $$U = \bigcup \{\mu(a, X\setminus B)\mid a \in A\}, V = \bigcup \{\mu(b, X\setminus A)\mid b \in B\}$$
and note that $A \subseteq U$, $B \subseteq V$ and these sets are open. They are also disjoint, for if there not, there must be some $a_0 \in A$ and some $b_0 \in B$ such that $\mu(a_0, X\setminus B) \cap \mu(b_0, X\setminus A) \neq \emptyset$ and by property 2 we'd then have $a_0 \in x \setminus A$ or $b_0 \in X\setminus B$, both of which are absurd. So that completes the proof.
Useful fact:
The proof is an easy exercise in the definitions.
Some examples: Any metric space is monotonically normal, we can define $\mu$ on a base by $\mu(x, B(x,r)) = B(x, \frac{r}{2})$, and the triangle inequality takes care of defining property 2.
The Sorgenfrey line is monotonically normal: define $\mu(x, [a,b)) = [x,b)$ on base elements, and checking property 2 is a simple matter of case checking.
So we then have "for free" that metric spaces are $T_5$ and so is the Sorgenfrey line. In fact all ordered spaces are monotonically normal too, this has a longer proof that I won't give here now.
But back to the Michael Line, or Scattered line: this is a special case of a more general construction: if $(X, \mathcal{T}_X)$ is a space and $A \subseteq X$ a subspace of $X$, then $X_A$ is the space with point set $X$ and topology generated by $\mathcal{T}_X \cup \mathscr{P}(X\setminus A)$; in effect: $A$ keeps its original subspace topology and all points outside of $A$ become isolated points.
The Michael line is just $\Bbb R_{\Bbb Q}$, where $\Bbb R$ has its usual topology.
Proposition
Proof: $T_1$ is trivial, as we have a finer topology. Let $\mu$ be the operator on $X$. Then a $\mu$ operator on the generating set of $X_A$ is defined by $\mu'(x,O)=\mu(x,O)$ if $O \in \mathcal{T}_X$ and $\mu'(x,B)=\{x\}$ if $B \subseteq X\setminus A$. This works, as is quite easy to check.
Corollary
Many spaces with good high separation properties have it for the reason that they're monotonically normal. No longer a very fashionable notion/property, but I quite like it.
Hope this helps someone.