# Finite and Countable Products of the Michael Line

Consider the real number line $\mathbb{R}$ with a topology stronger than the Euclidean topology such that the irrational numbers are isolated and the rational numbers retain their Euclidean open neighborhoods. When the real number line is endowed with this topology, the resulting topological space is called the Michael line and is denoted by $\mathbb{M}$. It is well known that $\mathbb{M} \times \mathbb{P}$ is not normal where $\mathbb{P}$ is the space of irrational numbers with the Euclidean topology. This and other basic results about the Michael line are discussed in the post Michael Line Basics. In this post, we show that $\mathbb{M}^n$ is paracompact for any positive integer $n$ and that $\mathbb{M}^\omega$ (the product of countably and infinitely many copies of $\mathbb{M}$) is not normal. Thus the Michael line is an example demonstrating that even when paracompactness is preserved by taking finite products, it can be destroyed by taking infinite product.

The results discussed in this post are from a paper by E. Michael (Example 1.1 in [2]). This paper had been discussed previously in this blog (see Two footnotes in a paper of E. Michael).

As discussed before, 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 the topology for the Michael line $\mathbb{M}$.

$\mathcal{B}=\tau \cup \left\{\left\{ x \right\}: x \in \mathbb{P}\right\}$

Other basic results about the Michael line are discussed in Michael Line Basics.

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Paracompactness

A space $X$ is paracompact if every open cover of $X$ has a locally finite open refinement. In proving $\mathbb{M}^n$ is paracompact, we need two basic results about paracompactness. The proof of Theorem 1 can be found in [1] (Theorem 5.1.11 in page 302) or in [3] (Theorem 20.7 in page 146). We prove Theorem 2.

Theorem 1
Let $X$ be a regular space. Then $X$ is paracompact if and only if every open cover of $X$ has an open $\sigma$-locally finite refinement, i.e., the following holds:

every open cover $\mathcal{U}$ of $X$ has an open refinement $\bigcup \limits_{j=1}^\infty \mathcal{V}_j$ such that each $\mathcal{V}_j$ is a locally finite collection of open subsets of $X$.

Theorem 2
Let $X$ be a regular space. Then $X$ is paracompact if and only if the following hold:

For each open cover $\left\{U_t: t \in T \right\}$ of $X$, there exists a locally finite open cover $\left\{V_t: t \in T \right\}$ such that $\overline{V_t} \subset U_t$ for each $t \in T$.

Proof of Theorem 2
The direction $\Longleftarrow$ is clear.

$\Longrightarrow$ Let $X$ be paracompact. Let $\mathcal{U}=\left\{U_t: t \in T \right\}$ be an open cover of $X$. By regularity, there is an open cover $\mathcal{W}$ of $X$ such that $\left\{\overline{W}: W \in \mathcal{W} \right\}$ refines $\mathcal{U}$. Since $X$ is paracompact, $\mathcal{W}$ has an open locally finite refinement $\mathcal{H}=\left\{H_a: a \in A \right\}$.

We now tie $\mathcal{H}$ to the original open cover $\mathcal{U}$. For each $a \in A$, choose $f(a) \in T$ such that $\overline{H_a} \subset U_{f(a)}$. Now, we go the opposite direction, i.e., for each $t \in T$, consider all $a \in A$ such that $\overline{H_a} \subset U_{f(a)}=U_t$. For each $t \in T$, let $V_t$ be defined by:

$V_t=\bigcup \left\{H_a: a \in A \text { and } \overline{H_a} \subset U_{f(a)}=U_t \right\}=\bigcup \limits_{f(a)=t} H_a$

Each $V_t$ is open since it is a union of open sets. Since $\mathcal{H}$ is locally finite, any subcollection of $\mathcal{H}$ is closure preserving. We have:

$\overline{V_t}=\bigcup \left\{\overline{H_a}: a \in A \text { and } \overline{H_a} \subset U_{f(a)}=U_t \right\}=\bigcup \limits_{f(a)=t} \overline{H_a}$

Thus we have $\overline{V_t} \subset U_t$ for all $t \in T$. Since $\mathcal{H}$ is locally finite, $\left\{V_t: t \in T \right\}$ is locally finite. Furthermore, $\left\{V_t: t \in T \right\}$ is clearly a cover of $X$. Thus Theorem 2 is established. $\blacksquare$

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Finite Products

What makes the Michael line finitely productive for paracompactness is that all but countably many points in $\mathbb{M}$ are isolated. The paracompactness of the finite products of the Michael line follows from Theorem 4 (see Corollary 5 below). Lemma 3 is used in proving Theorem 4.

Lemma 3
Let $X$ be a space such that all but countably many points of $X$ are isolated. Let $A$ be the set of all isolated points of $X$. Then for each $n=2,3,4,\cdots$, $X-A^n$ can be expressed as the following:

$X-A^n=\bigcup \limits_{k=1}^\infty Y_k$ satisfying the following:

• For each $k$, $Y_k$ is homeomorphic to $X^{n-1}$.
• For each $k$, there exists a continuous map $F_k:X^n \rightarrow Y_k$ such that $F_k \upharpoonright Y_k$ is the indentity map.

Proof of Lemma 3
Note that $X-A$ is countable. Fix $n \in \left\{2,3,4,\cdots \right\}$. For each $x \in X^n$, express $x=(x_1,x_2,\cdots,x_n)$. For each $i and for each $a \in X-A$, let $Y_{i,a}=\left\{x \in X^n: x_i=a \right\}$ (the $i^{th}$ coordinate is fixed and the other $n-1$ coordinates are free to vary). There are only countably many such $Y_{i,a}$. Clearly $X-A^n$ is the union of all $Y_{i,a}$. Furthermore, each $Y_{i,a}$ is homeomorphic to $X^{n-1}$.

Define $F_{i,a}:X^n \rightarrow Y_{i,a}$ by mapping each $(x_1,x_2,\cdots,x_i,\cdots,x_n)$ to $(x_1,x_2,\cdots,a,\cdots,x_n)$. In other words, the $i^{th}$ coordinate of each point is mapped to the fixed point $a$. This is a continuous map since it is a projection map. It is clear that when this map is restricted to $Y_{i,a}$, it is the identity map.

When we order all $Y_{i,a}$ in a sequence $Y_1,Y_2,Y_3,\cdots$, the lemma is established. $\blacksquare$

Theorem 4
Let $X$ be a regular space such that all but countably many points of $X$ are isolated. Then $X^n$ is paracompact for each $n=1,2,3,\cdots$.

Proof of Theorem 4
We prove $X^n$ is paracompact by induction on $n$. Let $A$ be the set of all isolated points of $X$. Let $B=X-A$.

First we show $X$ is paracompact. Let $\mathcal{U}$ be an open cover of $X$. Enumerate $B$ by $\left\{b_1,b_2,b_3,\cdots \right\}$. For each $i$, choose $U_i \in \mathcal{U}$ with $b_i \in U_i$. Let $\mathcal{U}_i=\left\{U_i \right\}$. Let $\mathcal{A}$ be the set of all $\left\{ x \right\}$ where $x \notin \bigcup \limits_{i=1}^\infty U_i$. Then $\mathcal{A} \cup \mathcal{U}_1 \cup \mathcal{U}_2 \cup \mathcal{U}_3 \cup \cdots$ is an open $\sigma$-locally finite refinement of $\mathcal{U}$. By Theorem 1, $X$ is paracompact.

Suppose that $X^{n-1}$ is paracompact where $n \ge 2$. Let $\mathcal{U}=\left\{U_t: t \in T \right\}$ be an open cover of $X^n$. By Lemma 3, there exist $Y_1,Y_2,Y_3,\cdots$, all subspaces of $X^n$, such that:

$X-A^n=\bigcup \limits_{k=1}^\infty Y_k$ satisfying the following:

• For each $k$, $Y_k$ is homeomorphic to $X^{n-1}$.
• For each $k$, there exists a continuous map $F_k:X^n \rightarrow Y_k$ such that $F_k \upharpoonright Y_k$ is the indentity map.

Fix $k$ where $k=1,2,3,\cdots$. Note that $\left\{U_t \cap Y_k: t \in T \right\}$ is an open cover of $Y_k$. Since each $Y_k$ is paracompact, using Theorem 2, we can find a locally finite open refinement $\left\{V_t: t \in T \right\}$ (open in $Y_k$) of $\left\{U_t \cap Y_k: t \in T \right\}$ such that $V_t \subset U_t \cap Y_k$ for each $t \in T$. For each $t$, let $W_{k,t}=F_k^{-1}(V_t) \cap U_t$.

Then $\left\{W_{k,t}: t \in T \right\}$ is a locally finite collection of open subsets of $X^n$ covering $Y_k$. Since the map $F_k$ is identity on $Y_k$, $V_t \subset F_k^{-1}(V_t)$. Thus $\left\{W_{k,t}: t \in T \right\}$ is a cover of $Y_k$. To see that it is locally finite, let $z \in X^n$. We have $F_k(z) \in Y_k$. There exists $V$ (open in $Y_k$) such that $F_k(z) \in V$ and $V$ only meets finitely many $V_t$, say, $V_{t_1},V_{t_2},\cdots,V_{t_m}$. Consider the following the open sets:

$W_{k,t_1}=F_k^{-1}(V_{t_1}) \cap U_{t_1}$
$W_{k,t_2}=F_k^{-1}(V_{t_2}) \cap U_{t_2}$

$\cdot$
$\cdot$
$\cdot$

$W_{k,t_m}=F_k^{-1}(V_{t_m}) \cap U_{t_m}$

$F_k^{-1}(V)$ is an open set containing $z$. It follows that the open sets $W_{k,t}$ that $F_k^{-1}(V)$ can meet are limited to ones listed above. For any $s \in T$ where $s \notin \left\{t_1,t_2,\cdots,t_m \right\}$, $V_s \cap V=\varnothing$. Thus $F_k^{-1}(V) \cap F_k^{-1}(V_s)=\varnothing$ and $F_k^{-1}(V) \cap W_s=\varnothing$. Thus $\left\{W_{k,t}: t \in T \right\}$ is locally finite in $X^n$.

For each $k=1,2,3,\cdots$, let $\mathcal{W}_k$ be $\left\{W_{k,t}: t \in T \right\}$, which is an open locally finite collection covering $Y_k$ (as shown above). All together $\bigcup \limits_{k=1}^\infty \mathcal{W}_k$ is an open $\sigma$-locally finite collection covering $X^n-A^n$. Let $\mathcal{A}$ be the set of all $\left\{ x \right\}$ where $x \in A^n$. Then $\mathcal{A} \cup \bigcup \limits_{k=1}^\infty \mathcal{W}_k$ is an open $\sigma$-locally finite refinement of $\mathcal{U}$. By Theorem 1, $X^n$ is paracompact. $\blacksquare$

Corollary 5
For each $n=1,2,3,\cdots$, $\mathbb{M}^n$ is paracompact.

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Infinite Products

Let $\mathbb{P}$ be the space of the irrational numbers with the Euclidean topology. Let $\omega$ be the set of all nonnegative integers. We now show that $\mathbb{M}^\omega$, the product of countably and infinitely many copies of the Michael line, is not normal. Before doing that, we point out that when $\omega$ is considered a discrete space, $\omega^\omega$, the product of countably and infinitely many copies of $\omega$, is homeomorphic to $\mathbb{P}$ (Thinking about the Space of Irrationals Topologically).

Let $D$ be a countably infinite subset of the Michael line $\mathbb{M}$ such that $D$ is closed and discrete. As discussed above, $D^\omega$ is a homeomorphic copy of $\mathbb{P}$. Furthermore $D^\omega$ is a closed subset of $\mathbb{M}^\omega$. Thus $\mathbb{M} \times \mathbb{M}^\omega$ contains $\mathbb{M} \times D^\omega \cong \mathbb{M} \times \mathbb{P}$ as a closed subspace. Since $\mathbb{M} \times \mathbb{P}$ is not normal, $\mathbb{M} \times \mathbb{M}^\omega$ is not normal. On the other hand, $\mathbb{M}^\omega$ is homeomorphic to $\mathbb{M} \times \mathbb{M}^\omega$. Thus $\mathbb{M}^\omega$ is not normal. $\blacksquare$

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Reference

1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Michael, E., Paracompactness and the Lindelof property in Finite and Countable Cartesian Products, Compositio Math., 23, 1971, 199-214.
3. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

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$\copyright \ \ 2012$