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):

which ones does $X$ satisfy?

This should depend on the cardinality of $S$.

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 $\mathcal V'$.

What about when $S$ is infinite?


1 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 $(-1/n,0)\cup(0,1/n)$.

  • 1
    $\begingroup$ Couldn't you replace 'metacompact' with 'metaLindelöf' (or even 'submetaLindelöf') in the last paragraph? $\endgroup$
    – Tyrone
    Commented Aug 21, 2023 at 10:02
  • $\begingroup$ Sure, you could. $\endgroup$ Commented Aug 21, 2023 at 14:49

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