# Paracompactness properties of the line with multiple origins

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?

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