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Wikipedia defines a $T_1$ space as: [1]

a T1 space is a topological space in which, for every pair of distinct points, each has a neighborhood not containing the other point.

Wikipedia defines a Hausdorff space the same way [2]:

a Hausdorff space, separated space or T2 space is a topological space where for any two distinct points there exist neighbourhoods of each which are disjoint from each other

The Munkres Topology textbook agrees with the above Hausdorff definition but disagrees on the $T_1$ definition. Munkres says that a $T_1$ space is one where every finite point set is closed, which is a weaker condition than the Hausdorff condition. Every Hausdorff space is a $T_1$ space, but not vice versa.

The real line $\mathbb{R}$ in the finite complement toplogy is not a Hausdorff space, but is a $T_1$ space in that every finite point set is closed.

[1] https://en.wikipedia.org/wiki/T1_space

[2] https://en.wikipedia.org/wiki/Hausdorff_space

clay
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    Wikipedia mentions Munkres's definition as equivalent (see https://en.wikipedia.org/wiki/T1_space#Properties). I'm sure the question of how to prove the equivalence has been asked here many times. – Eric Wofsey Feb 15 '20 at 21:05
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    https://math.stackexchange.com/questions/1830929/in-t1-space-all-singleton-sets-are-closed – Moishe Kohan Feb 15 '20 at 21:14
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    You should clarify your question: It appears that you are asking about equivalence of the two definitions of T1: One in Munkres and the other in Wikipedia, while the body of your question discusses something else, namely, the notion of a Hausdorff space and the body of the question contains no questions at all (only definitions and statements). – Moishe Kohan Feb 15 '20 at 21:20
  • To maybe aid your intuition, I like to think about it like this: T1 = "any two points are topologically distinct." T2 = "any two points are topologically separated." To see why T1 is the same as "singletons are closed," think about the closure. If points are topologically distinct, then no points are close to any other points, i.e. the closure of a singleton is itself. To see why T1 is different from T2, simply consider the topology on ${1, 2, 3}$ with open sets $\emptyset, {1, 3}, {2, 3}, {3}, {1, 2, 3}$. – Charles Hudgins Dec 25 '21 at 16:32

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The definition on Wikipedia is correct, and it's different from being Hausdorff.

Let's look at an example, the cofinite topology on $\Bbb N$, where the open sets are the subsets of $\Bbb N$ whose complement is finite. Given $m,n\in\Bbb N$ we can find a neighbourhood of $m$ not containing $n$, such as $\Bbb N\setminus\{n\}$, and a neighbourhood of $n$ not containing $m$, such as $\Bbb N\setminus\{m\}$. This space however is not Hausdorff since every two nonempty open sets have nonempty intersection, this example should convince you that the two notions are different.

The definition on Wikipedia for $T_1$ is equivalent to that of Munkres, I suggest you convince yourself of the equivalence as an exercise, but it has been asked on this website before if you want to look it up.

Alessandro Codenotti
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Just to be clear, $T_1$ and $T_2$ are not the same. The question is then whether two definitions of $T_1$ agree, and indeed they do.

Claim: For a space $X$, the following are equivalent.

  1. Given distinct points $x,y\in X$, there exist open sets $U$ and $V$ such that $x\in U$, $y\in V$, $y\notin U$, and $x\notin V$.

  2. Any finite set is closed.

Proof:

$(1\implies 2):$ By induction it suffices to show a singleton is closed. Fix $x\in X$. For any $y\neq x$, there exists an open set $U_y$ containing $y$ but not $x$. Thus $X\setminus U_y$ is closed and contains $x$ but not $y$, so $\{x\}=\bigcap_{y\in X}(X\setminus U_y)$ is closed.

$(2\implies 1):$ Take $x\neq y$. Then $X\setminus \{y\}$ and $X\setminus\{x\}$ have the desired properties.

Community
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pancini
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  • No induction is necessary. Since in any topological space a finite intersection of open sets is open, the same can be said for finite unions of closed sets, by taking complements. –  Feb 15 '20 at 22:11
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    That's what I meant by induction. – pancini Feb 15 '20 at 22:12
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It doesn't sound like it's the same definition. Wikipedia says a $T_1$ space is a topological space where for every pair of distinct points, each of those points has a neighborhood not containing the other point itself - that is, not that other point's neighborhood. That is, this definition for $T_1$ can hold even if the two points' respective neighborhoods were not disjoint.

Que
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The definitions are different. There's a difference between having disjoint nbhds and each point having a nbhd not containing the other.

$T_1$ is equivalent to points being closed, which is equivalent to finite sets being closed.