Paracompactness of the Everywhere doubled line

Question

The Everywhere doubled line is an interesting example of a non-Hausdorff manifold that is homogeneous. It is described in section 3 of [BG], similar to the line with two origins but with each point "doubled". The article has some nice pictures.

The space can be described as the union of two real lines $X=A\sqcup B$ with $A = \mathbb R\times\{0\}$ and $B=\mathbb R\times\{1\}$. A set $U\subseteq X$ is open exactly when $U\cap A$ is open in $A$ with the usual Euclidean topology and for each $x=(r,1)\in U\cap B$, the set $U$ contains a deleted neighborhood of $(r,0)$ in $A$. It is a special case of the general construction here.

I am wondering about the paracompactness and related properties of $X$ (similar to this question).

For one thing, the space is separable ($\mathbb Q\times\{0\}$ is dense in $X$), but is not Lindelöf, as the cover of $X$ consisting of the open sets $(\mathbb R\times\{0\})\cup\{(r,1)\}$ for $r\in\mathbb R$ has no countable subcover. So the space is not metacompact, since metacompact separable spaces are Lindelöf (proof here for example). And therefore the space is not paracompact either.

I have convinced myself that

The Everywhere doubled line is not countably metacompact.

(and hence not countably paracompact either).

Can anyone provide a proof?

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[BG] M. Baillif & A. Gabard, Manifolds: Hausdorffness versus homogeneity, Proc. Amer. Math. Soc. 136 (2008), 1105-1111

Answer 1

Proof that $X$ is not countably metacompact:

Partition the reals into countably many dense sets: $\mathbb R=\bigcup_{n=1}^{\infty}D_n$. Each of the sets $A_n=(\mathbb R\times\{0\})\cup(D_n\times\{1\})$ is open in $X$ and the $A_n$ form a countable open cover $\mathcal U$ of $X$. I claim that this cover does not admit a point finite open refinement.

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.

So suppose $\mathcal V$ is an open refinement of $\mathcal U$. Take $a_1\in D_1$. The point $x_1=(a_1,1)\in X$ is in some open set $V_1\in\mathcal V$. Since $V_1$ is open, it contains some nonempty open interval $W_1=(a_1,b_1)\subseteq\mathbb R$ with left endpoint equal to $a_1$. Now $D_2$ is dense in $\mathbb R$, so we can choose $a_2\in D_2\cap(a_1,b_1)$. The point $x_2=(a_2,1)\in X$ is in some open set $V_2\in\mathcal V$. And since the sets $D_n$ are pairwise disjoint, the points $x_1$ and $x_2$ cannot belong to a common $U\in\mathcal U$, and the same in the refinement $\mathcal V$. So necessarily $V_1\ne V_2$. Since $V_2$ is open, it contains some nonempty open interval $W_2=(b_2,a_2)\subseteq(a_1,b_1)\subseteq\mathbb R$ with right endpoint equal to $a_2$.

Continuing in this way, alternating between left endpoint and right endpoint for the elements $a_n$, we get a sequence of open intervals $W_n\subseteq\mathbb R$ with $W_n\subseteq V_n\in\mathcal V$ and all the $V_n$ pairwise distinct. And due to the alternating choices of left and right endpoints, the intersection of all the $W_n$ is nonempty (for example, check that the odd entries $a_{2n+1}$ are strictly increasing and the even entries $a_{2n}$ are strictly decreasing, etc.). Now any $x\in X$ in this common intersection belongs to all the $V_n$, which shows that the cover $\mathcal V$ is not point finite.

Answer 2

@PatrickR has a solid proof showing this space fails countable metacompactness. Here's note to go with it.

I was a little shook when I realized this space was indeed homogeneous. After all, $\mathbb R\times\{0\}$ is a copy of the real line, but $\mathbb R\times\{1\}$ is discrete; these points feel different. But consider the points $\langle x,0\rangle,\langle x,1\rangle$; the function swapping only these two points is a homeomorphism. Then given any two points $\langle x,i\rangle,\langle y,j\rangle$ with $x\not=y$, let $f$ be a homeomorphism where $\langle x,i\rangle\mapsto_f\langle x,0\rangle$, let $g$ be the homeomorphism $\langle z,k\rangle\mapsto_g\langle z+y-x,k\rangle$, and let $h$ be a homeomorphism where $\langle y,0\rangle\mapsto_h\langle y,j\rangle$; then $\theta=h\circ g\circ f$ is a homeomorphism where $\langle x,i\rangle \mapsto_\theta \langle y,j\rangle$.