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I was freshing up on some topology, and this text I'm reading mentions T1 does not imply Hausdorff. A few counter-examples are readily available, like the natural numbers under the co-finite topology.

But what if we place a restriction on the space to also be compact, the text doesn't mention anything about that and I can't come up with any examples of spaces that are compact and T1 but not Hausdorff. To state it otherwise, I'm looking for a space that is T1 but not normal(=compact and Hausdorff).

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  • $\begingroup$ I think a compactification of the line with two origins works? $\endgroup$ – B. Mehta May 22 '17 at 16:41
  • $\begingroup$ Or just the same construction but starting with $[-1, 1]$ instead of $\mathbb{R}$ $\endgroup$ – B. Mehta May 22 '17 at 16:42
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    $\begingroup$ Also, the cofinite topology on an infinite set works. $\endgroup$ – B. Mehta May 22 '17 at 16:43
  • $\begingroup$ normal is not the same as compact plus Hausdorff, it's implied by it. For a compact $T_1$ space $X$ it is true that : $X$ is Hausdorff iff $X$ is normal, which is probably what you meant to say. $\endgroup$ – Henno Brandsma May 23 '17 at 6:53
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You've already given an example: the natural numbers (or any infinite set, really) under the co-finite topology. Given any open cover, fixing a single (non-empty) element of the cover yields an open set that has all but finitely-many of the natural numbers as elements. Thus, only finitely-many more elements of the cover are needed, forming a finite subcover.

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The affine space $\Bbb C^n$ endowed with the Zariski topology: $C$ is closed if and only if there is a family of multivariate polynomials $\mathcal F_C\subseteq \Bbb C[x_1,\cdots,x_n]$ such that $$C=\{x\in\Bbb C^n\,:\,\forall f\in\mathcal F_C,\ f(x)=0\}$$

These topologies have the following notable properties:

  • they are T1;
  • any two non-empty open sets have non-empty intersection (irreducibility): thus the topology is not T2, and on a side note all open subsets are connected;
  • for any non empty family $\mathfrak C$ of closed sets, there is $C\in\mathfrak C$ which is maximal with respect to inclusion "$\subseteq$" (noetherianity): thus every subspace is compact.

The proofs of these facts are not difficult, but they use a couple of lemmas of commutative algebra which might make the exposition a bit long.

An easier special case of this is when $n=1$, in which case the Zariski topology is just the cofinite topology on $\Bbb C$ (the topology where a set is open if and only if its complement is either finite or the whole space).

Addendum: I assumed here that your definition of "compact topological space" is:

Cpt: A topological space such that any open cover $U$ admits a finite subcover.

However, a considerable number of authors (for instance, Bourbaki), call "quasi-compact" a topological space which satisfies Cpt and "compact" a T2 quasi-compact space. In this case, though, your question would be trivial.

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  • $\begingroup$ Engelking (General Topology) calls a compact $T_2$ space a compact space, and a space in which every open cover has a finite subcover but is not $T_2$ is called pseudo-compact. $\endgroup$ – DanielWainfleet May 23 '17 at 10:12
  • $\begingroup$ @DanielWainfleet Curiously, encyclopediaofmath cites that book as a reference for "pseudo-compact = completely regular + every continuous $f:X\to\Bbb R$ is bounded". $\endgroup$ – user228113 May 23 '17 at 10:36
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    $\begingroup$ Thank you. I will have to re-check my copy for his def'n. $\endgroup$ – DanielWainfleet May 23 '17 at 20:43
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Let $T_X$ be a compact Hausdorff topology on an infinite set $X$, with no isolated points. So every non-empty member of $T_X$ is an infinite set. Let $Y=X\times \{1,2\}.$ $$\text {Let} \quad B= \{\;( t\times \{1,2\} )\; \backslash A:t\in T_X \land A \text { is finite} \}.$$ Then $B$ is a base for a topology $T_Y$ on $Y.$

(i). For each $p\in Y$ we have $Y$ \ $\{p\}=(X\times \{1,2\})$ \ $\{p\}\in B\subset T_Y.$ So $T_Y$ is a $T_1$ topology.

(ii). For $x\in X$ and $i\in \{1,2\},$ if $(x,i)\in U_i\in T_Y,$ then $(x,i)\in (t_i\times \{1,2\})$ \ $A_i\subset U_i$ for some $t_i\in T_X$ and some finite $A_i.$

Then $U_1\cap U_2\supset (t_1\cap t_2)$ \ $(A_1\cup A_2)\ne \phi .$ So $(x,1)$ and $(x,2)$ do not have disjoint nbhds: $T_Y$ is not a Hausdorff topology.

(iii). For $i\in \{1,2\}$ the subspace $X\times \{i\} $ is homeomorphic to $X$ so it is compact. Now $Y$ is the union of the two compact subspaces $X\times \{1\},\;X\times \{2\}$ so $Y$ is also compact.

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