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The Everywhere doubled line is an interesting example of a non-Hausdorff manifold that is homogeneous. See this question for some picture.

The space can be described as the union of two real lines $X=(\mathbb R\times\{0\})\cup(\mathbb R\times\{1\})$. For $r\in\mathbb R$, nbhds of the point $x=(r,0)$ are the sets containing a Euclidean nbhd in $\mathbb R\times\{0\}$. Nbhds of the point $x=(r,1)$ are the sets containing $x$ and a "deleted nbhd" of $(r,0)$ in $\mathbb R\times\{0\}$.

Paracompactness of the Everywhere doubled line showed that the space is not countably metacompact (hence not paracompact or metacompact). One can ask about other weaker properties related to paracompactness.

Question: Is the Everywhere doubled line submetacompact or submeta-Lindelöf?

Here are the definitions:

  • metacompact: every open cover has a point finite open refinement.
  • meta-Lindelöf: every open cover has a point countable open refinement.
  • submetacompact (= $\theta$-refinable): for every open cover $\mathscr U$, there is a sequence of open covers $(\mathscr U_n)_n$, each a refinement of $\mathscr U$, such that for each $x\in X$ there is some $n$ with $\{U\in\mathscr U_n:x\in U\}$ finite.
  • submeta-Lindelöf (= $\delta\theta$-refinable): for every open cover $\mathscr U$, there is a sequence of open covers $(\mathscr U_n)_n$, each a refinement of $\mathscr U$, such that for each $x\in X$ there is some $n$ with $\{U\in\mathscr U_n:x\in U\}$ countable.

with the evident implications:

$$\text{metacompact}\implies\text{meta-Lindelöf}\implies\text{submeta-Lindelöf}$$ $$\text{metacompact}\implies\text{submetacompact}\implies\text{submeta-Lindelöf}$$

[This is a self-answered question.]

PatrickR
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1 Answers1

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Answer: $X$ is not submeta-Lindelöf (and hence not submetacompact and not meta-Lindelöf).

To simplify the notation, I'll denote the "bottom" line $\mathbb R\times\{0\}$ just by $\mathbb R$, so $X=\mathbb R\cup(\mathbb R\times\{1\})$, assuming that the two lines are disjoint. Also, for $r\in\mathbb R$ and $\varepsilon>0$, write $B(r,\varepsilon)=(r-\varepsilon,r+\varepsilon)$ for the ball and $B'(r,\varepsilon)=B(r,\varepsilon)\setminus\{r\}$ for the ball without its center.

To show $X$ is not submeta-Lindelöf , consider the open cover $\mathscr U$ consisting of all sets of the form $(\mathbb R\times\{0\})\cup\{x\}$ with $x$ a point on the top line.

Step 1: Suppose $\mathscr V$ is an open refinement of $\mathscr U$. Each point $x=(r,1)$ on the top line belongs to some $V_x\in\mathscr V$ contained in some member of $\mathscr U$. And because each point on the top line only belongs to one member of $\mathscr U$, all these $V_x$ are distinct. By definition of the topology of $X$, there is some $\varepsilon_r>0$ such that $B'(r,\varepsilon_r)\subseteq V_x$. (Note that $B'(r,\varepsilon_r)$ determines $V_x$ and, for $t\ne r$ in $\mathbb R$, $t\in B'(r,\varepsilon_r)$ iff $t\in B(r,\varepsilon_r)$.)

Claim: $S=\{t\in\mathbb R: t\in B(r,\varepsilon_r)\text{ for uncountably many }r\}$ is open and dense in $\mathbb R$.

To show it is dense in $\mathbb R$, consider an interval $(a,b)$ with $a<b$. Since that interval is uncountable, there is some integer $m\ge 1$ such that $T=\{r\in(a,b):\varepsilon_r>\frac1m\}$ is uncountable. Some element $t\in T$ is a condensation point of $T$ (see for example here). Therefore, the interval $(t-\frac1m,t+\frac1m)$ contains uncountably many elements $r\in T$; and for each such $r$ in that interval, $\varepsilon_r>\frac1m$, so that $t\in B(r,\varepsilon_r)$. That is, $t\in S$ as required.

To show $S$ is open in $\mathbb R$, let $t\in S$. If $t$ belongs to the open ball $B(r,\varepsilon_r)$, then $B(t,1/m)\subseteq B(r,\varepsilon_r)$ for some integer $m$. So if $t$ belongs to uncountably many $B(r,\varepsilon_r)$, there is some $m$ for which $B(t,1/m)$ is contained in uncountably many $B(r,\varepsilon_r)$. And thus $B(t,1/m)\subseteq S$.

From the claim, it follows that $\{t\in\mathbb R: t\in V_x\text{ for uncountably many }x=(r,1)\}$ contains some open dense subset of $\mathbb R$.

Step 2: Let $\mathscr U_1,\mathscr U_2,\dots$ be a sequence of open covers of $X$, each a refinement of $\mathscr U$. By Step 1, there is an open dense set $O_n\subseteq\mathbb R$ with each $t\in O_n$ contained in uncountably many members of $\mathscr U_n$. By the Baire category theorem there is an element $t\in\mathbb R$ that belongs to uncountably many members of $\mathscr U_n$ for all $n$. This shows that $X$ is not submeta-Lindelöf.

PatrickR
  • 7,165