Working with non-regular topological semigroups, my collegue Oleg Gutik discovered a special space $H$ which we named Gutik's hedgehog. It is homeomorphic to the space $$H:=\{(0,0)\}\cup\{(\tfrac1n,0):n\in\mathbb N\}\cup\{(\tfrac1n,\tfrac1{nm}):n,m\in\mathbb N\},$$ endowed with the topology $\tau$ consisting of sets $U\subset H$ satisfying the following two conditions:

(1) if $(\frac1n,0)\in U$ for some $n\in\mathbb N$, then there exists $m\in\mathbb N$ such that $(\frac1n,\frac1{nk})\in U$ for all $k\ge m$;

(2) if $(0,0)\in U$, then there exists $m\in\mathbb N$ such that $(\frac1n,\frac1{nk})\in U$ for all $n\ge m$ and all $k\in\mathbb N$.

It turns out that Gutik's hedgehog is a test space for regularity in the class of first-countable Hausdorff spaces.

Theorem. A first-countable Hausdorff space is regular if and only if it contains no topological copies of the Gutik hedgehog.

Because of this fundamental role in testing regularity, I admit that Gutik's hedgehog is known in topology under some different name. I would be grateful for any information in this respect.

Remark 1. The Gutik's hedgehog resembles (but is not equal to) the non-regular space of Smirnov, see Example 64 in "Counterexamples in Topology".

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    $\begingroup$ Whatever other name there might be, I prefer "Gutik's hedgehog". $\endgroup$ – Nik Weaver Mar 28 '18 at 11:24
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    $\begingroup$ I also like the hedgehog name. But as a topological space, what is the difference between the Gutik hedgehog and $\omega^2+1$? Unless I've misunderstood, it seems that you have a convergent sequence of convergent sequences, just like $\omega^2+1$. $\endgroup$ – Joel David Hamkins Mar 28 '18 at 12:25
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    $\begingroup$ Hmm, it looks more like a porcupine to me. $\endgroup$ – Jeremy Rickard Mar 28 '18 at 12:46
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    $\begingroup$ My thought was that a hedgehog, facing to the left, has a maximum around halfway. But a porcupine is closer to monotone increasing. $\endgroup$ – Jeremy Rickard Mar 28 '18 at 12:59
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    $\begingroup$ Difference between hedgehog and porcupine: youtube.com/watch?v=jzKCu9dUcOw. $\endgroup$ – Joel David Hamkins Mar 28 '18 at 13:04

Searching Steen & Seebach, Counterexamples in Topology for "Hausdorff, separable, not regular" I found this example. Is that it?

This is Example 79, page 97.

Irregular Lattice Topology
Let $A = \{(i,k) : 0 < i,k \in \mathbb{N}\}$, $B = \{(i,0) : i \geq 0\}$ and $X = A \cup B$. Declare each point of $A$ to be open. Let the set of all $U_n((i,0)) = \{(i,k) : k=0$ or $k \geq n\}$ form a local basis at any point with $i \neq 0$ and the set of all $V_n = \{(i,k) : i=k=0$ or $i,k\geq n\}$ be a local basis at $(0,0)$.

  • $\begingroup$ This is almost what is necessary, except for the definition of $V_n$. In the Gutik hedgehog $V_n=\{(i,k):i=k=0$ or $i\ge n\}$. The Gutik hedgehog is homeomorphic to the subspace $\{(0,0\}\cup\{(i,0):i\in\mathbb N\}\cup\{(i,k):0<i\le k\}$ of $A$. $\endgroup$ – Taras Banakh Mar 28 '18 at 13:02
  • $\begingroup$ So is this $X$ homeomorphic to $H$ anyway? $\endgroup$ – Gerald Edgar Mar 28 '18 at 13:11
  • $\begingroup$ Probably not: the regularization of Guik's hedgehog (or Gutik's porcupine) is compact but the regularization of $A$ is not. $\endgroup$ – Taras Banakh Mar 28 '18 at 13:16

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