Paracompactness properties of the line with multiple origins

Question

Let $X$ be the "line with multiple origins", obtained by taking a set $S$ with the discrete topology and taking the quotient space of $\mathbb R\times S$ by the equivalence relation that identifies $(x,\alpha)$ with $(x,\beta)$ whenever $x\ne 0$. This is a generalization of the classical line with two origins.

Alternatively, one can take the Euclidean line $\mathbb R$ and replace the origin $0$ with many origins $0_\alpha$ ($\alpha\in S$). Basic open nbhds of each origin $0_\alpha$​ are of the form $(U\setminus\{0\})\cup\{0_\alpha\}$ with $U$ a Euclidean open nbhd of $0$.

Of the following related topological properties (no Hausdorff assumption here):

• paracompact: every open cover has a locally finite open refinement.
• metacompact: every open cover has a point finite open refinement.
• countably paracompact: every countable open cover has a locally finite open refinement.
• countably metacompact: every countable open cover has a point finite open refinement.

which ones does $X$ satisfy?

This should depend on the cardinality of $S$.

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For $S$ finite, the space is paracompact (hence the other properties are also satisfied).

I don't know if this follows from general results, as $X$ is neither Hausdorff nor regular. But here is an ad-hoc proof for the line with two origins $0_1$ and $0_2$. The general case of $S$ finite is similar.

Suppose $\mathcal U$ is an open cover of $X$. The subspace $Y=(\mathbb R\setminus\{0\})\cup\{0_1\}$ is open in $X$ and homeomorphic to $\mathbb R$, hence paracompact. Intersecting every element of $\mathcal U$ with $Y$ gives an open cover of $Y$, and we can choose a locally finite open refinement $\mathcal V$ of that cover of $Y$. Taking any element $U\in\mathcal U$ containing $0_2$ and adjoining it to $\mathcal V$ provides an open refinement $\mathcal V'$ of $\mathcal U$.

Now to show that refinement is locally finite. For points of $Y$, that follows from the local finiteness of $\mathcal V$. And for the origin $0_2$, suppose $V_1$ is a nbhd of $0_1$ in $Y$ that witnesses local finiteness of $\mathcal V$ at $0_1$. The corresponding set obtained by replacing $0_1$ with $0_2$ in $V_1$ is a nbhd of $0_2$ that also meets only finitely many elements of $\mathcal V$, and the same of $\mathcal V'$.

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What about when $S$ is infinite?

Answer 1

$X$ is always countably metacompact (and thus metacompact if $S$ is countable since then $X$ is Lindelöf). Indeed, suppose $(U_n)$ is a countable open cover of $X$. Intersecting these sets with one of the copies of $\mathbb{R}$, we can get a point-finite refinement that covers that copy. Now add on the sets $U_n\cap (S\cup(-1/n,0)\cup(0,1/n))$ for each $n$ to get a cover of all of $X$. This cover will be point-finite except possibly at the points of $S$, if some element of $S$ is in infinitely many $U_n$'s. To fix this, just remove each element of $S$ from all but one of the sets used in the cover.

On the other hand, if $S$ is infinite, then $X$ is not countably paracompact. Indeed, if $S$ is countably infinite, then you can cover $X$ by the copies of $\mathbb{R}$ with each element of $S$, and this has no locally finite refinement since any refinement must contain a separate open set for each element of $S$ and these will all intersect any neighborhood of any of the origins. If $S$ is uncountable, you can do the same thing by taking countably many of the copies of $\mathbb{R}$ and then throwing the rest of the elements of $S$ into one of them.

Finally, if $S$ is uncountable, then $X$ is not metacompact. Again, consider the cover of $X$ by the copies of $\mathbb{R}$ for each element of $S$. A refinement must contain a separate open set for each element of $S$ and then since $S$ is uncountable there is some $n$ such that uncountably many of them contain $(-1/n,0)\cup(0,1/n)$.

Answer 2

(Follow-up to the answer by Eric Wofsey, regarding Tyrone's comment)

There are a few more related properties:

• 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:

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

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Which of these properties does $X$ (= line with multiple origins) satisfy?

(1) If $S$ is countable, $X$ is metacompact, and hence satisfies the three properties above.

(2) If $S$ is uncountable, $X$ is not meta-Lindelöf (same as Eric's argument for metacompact).

(3) $X$ is always submetacompact, and hence submeta-Lindelöf, independently of the number of origins.

Proof of (3): (with help from Tyrone) Let $\mathscr U$ be an open cover of $X$. The open subspace $Y=(-\infty,0)\cup(0,\infty)\subseteq X$ is metrizable, hence paracompact. Intersecting every element of $\mathscr U$ with $Y$ gives an open cover of $Y$, and we can choose a locally finite open refinement $\mathscr V$ of that cover of $Y$.

For each $\alpha\in S$ there is some $n_\alpha\ge 1$ such that $(-1/n_\alpha,0)\cup\{0_\alpha\}\cup(0,1/n_\alpha)$ is contained in some member of $\mathscr U$. Let $$V_{\alpha,n}=(-\frac{1}{n_\alpha+n},0)\cup\{0_\alpha\}\cup(0,\frac{1}{n_\alpha+n})$$

For each $n$, define $\mathscr U_n=\mathscr V\cup\{V_{\alpha,n}:\alpha\in S\},$ which is an open refinement of $\mathscr U$. The open covers $\mathscr U_n$ are as required.

Answer 3

I can't help but point out that PatrickR's proof of the submetacompactness of $X$ actually leads to a stronger conclusion.

Definition A space is subparacompact if each of its open covers has a $\sigma$-discrete closed refinement.

We have the following implications $$\text{paracompact regular}\implies\text{subparacompact}\implies\text{submetacompact}$$ The first of these follows from a result of E. Michael (note that each paracompact Hausdorff space is paracompact regular). The second of these implications follows from the following characterisation due to D. Burke.

Theorem (Burke) A space $X$ is subparacompact if and only if for each open cover $\mathcal{U}$ of $X$ there is a sequence $\{\mathcal{U}_n\}_{n\in\mathbb{N}}$ of open covers, each a refinement of $\mathcal{U}$, such that for each $x\in X$ there is $n\in\mathbb{N}$ such that $\{U\in\mathcal{U}_n\mid x\in U\}$ consists of exactly one element.

We have the following.

Proposition Let $S$ be a nonempty set and $X$ the line with $S$ origins. Then $X$ is subparacompact.

Proof (following PatrickR) Let $\mathcal{U}$ be an open cover of $X$. Then $Y=X\setminus S=(-\infty,0)\cup(0,\infty)$ is metrisable, so by Burke's Theorem there is a sequence of open covers $\{\mathcal{V}_n\}_{n\in\mathbb{N}}$ of $Y$ such that each refines $\mathcal{U}$ and for each point of $Y$ there is $n\in\mathbb{N}$ for which $\{V\in\mathcal{V}_n\mid y\in V\}$ has exactly one member. Since $Y$ is open in $X$, the members of each $\mathcal{V}_n$ are open in $X$.

Now, for each $\alpha\in S$ and $n\in\mathbb{N}$ let $V_{\alpha,n}$ be the open subset of $X$ defined by PatrickR. The three properties of these sets which we need are that $(i)$ each $V_{\alpha,n}$ includes in some member of $\mathcal{U}$, $(ii)$ $V_{\alpha,n}\cap S=\{\alpha\}$ for each $n\in\mathbb{N}$, and $(iii)$ if $x\in Y$ and $|x|>\frac{1}{N}$, then $x\not\in V_{\alpha,n}$ for any $\alpha\in S$ and $n\geq N$.

Finally, for $m,n\in\mathbb{N}$ let $\mathcal{U}_{m,n}=\mathcal{V}_n\cup\{V_{m,\alpha}\mid \alpha\in S\}$. Each $\mathcal{U}_{m,n}$ is an open cover of $X$ refining $\mathcal{U}$. If $\alpha\in S$, then $\alpha$ belongs to exactly one member of each $\mathcal{U}_{m,n}$. Furthermore, given $x\in Y\subseteq X$ choose $n\in\mathbb{N}$ such that $x$ belongs to exactly one member of $\mathcal{V}_n$. This $x$ belongs to exactly one member of $\mathcal{U}_{n,m}$ for any $m>\frac{1}{|x|}$.

By Burke's Theorem, we conclude that $X$ is subparacompact. $\quad\blacksquare$