# Is hedgehog of countable spininess separable space?

My question is about hedgehog space, which is defined in the same way as it can be found on Wikipedia.

It seems obvious that hedgehog of countable spininess is separable, because it is countable sum of an intervals with standard topology.

But there is theorem which says that every separable metric space is strongly paracompact (hypocompact). Let us cover hedgehog by a ball of radius $1/3$ centered at the origin and and by balls of radius $3/4$ centered at ends of spines. This cover is an open cover and I can't see how we can find star-finite refiniment (seems like set containing origin should intersect some set from every spine).

Where is flaw in my reasoning?

• maybe at the very least link to the wikipedia article Jan 13, 2018 at 13:48
• en.wikipedia.org/wiki/Hedgehog_space Strong paracompactness is defined as follows: for every open cover there is an open refinement such that every set from this refinement has nonempty intersection with only finitely many others Jan 13, 2018 at 13:50
• I'm confused. Does it, by definition, have the quotient topology given by identifying the origins of $\kappa$ disjoint copies of the unit interval? If so how can a metric ball be contained in the neighborhood of the origin given by intervals that shrink to the origin? Jan 13, 2018 at 14:20
• Lol, look at the talk page on the wiki article Jan 13, 2018 at 14:28
• Space which you described is not metrizable, because by your argument it does not satisfy first countability axiom. It is better to think about this as a slight modification of a British Rail metric. EDIT: Sorry, this exactly what you found on the talk page. Jan 13, 2018 at 14:29

Ok so you obviously have an open set $U$ of the refinement that contains the origin and which is contained in the original origin-centered ball $B$ or radius $1/3$. You're worried that $U$ has to intersect infinitely many refinement sets each of which is confined to its own spine, but thats not necessarily the case. A refinement set, $W$, that intersects $U$ could be contained in $B$, and so $W$ could be the disjoint union of pieces of the interiors of all the spines (subject to the star-finiteness condition). Thus $W$ could "bridge the gap" between $U$ and finitely many of the refinements of the $3/4$ balls. After that you do the same thing again: take a subset, $V$, of $B$ that bridges the gap between $W$ and a few more (finitely many) refinements of the $3/4$ balls, but also such that $V$ is disjoint from $U$ etc.
P.S. On a manifold, you can always take a refinement with an injective pairing, taking an element of the refinement to a superset element of the original cover. This would prohibit the above idea since $W$ would have to find its own superset element of the original cover ($B$ already being taken by $U$). I don't know if there's a name for this property and I don't remember what manifold axioms are used to get it -- it would be interesting and reassuring to see where it fails in the case of the hedgehog space.
Engelking has the theorem (5.3.10, due to Smirnov) that for regular $X$, $X$ is hypocompact iff every open cover has a star-countable refinement. Your cover is already star-countable, so would need no refinement.
From this it also follows that all Lindelöf regualr spaces are hypocompact, which implies the separable metric case. And the hedgehog $H(\kappa)$ with $\kappa$ spines has weight $\kappa$, where $\kappa$ is an infinite cardinal.