Help understanding $T_1$ & $T_2$ spaces.

Question

What are $T_1$ & $T_2$ spaces? I'm having trouble understanding what exactly are $T_1$ & $T_2$ spaces.

Answer 1

$T_{1}$ and $T_{2}$ are part of a larger collection of axioms, usually called the separation axioms. Most of these axioms assign some conditions (for a topological space) that separate points or closed sets from each other with open sets.

If $(X,\tau)$ is a topological space, then we say that it is

• $T_{1}$ if for every two points $x\neq y$ in $X$ there exists open sets $U_{x},U_{y}\subseteq X$ so that $x\in U_{x}$, $y\in U_{y}$, and $x\notin U_{y}$, $y\notin U_{x}$. In other words, if distinct points have open neighbourhoods that do not contain the other point. This condition can be shown to be equivalent with the condition of every singleton $\{x\}$ is a closed subset of $X$.

• $T_{2}$, or sometimes called Hausdorff, if for every two points $x\neq y$ in $X$ there exists disjoint open neighbourhoods, i.e. open sets $U_{x},U_{y}\subseteq X$ so that $x\in U_{x}$, $y\in U_{y}$ and $U_{x}\cap U_{y}=\emptyset$. This separation axiom provides many useful conditions, such as unique limits of converging sequences.

One can easily verify that $T_{2}$ is stronger than $T_{1}$, i.e. $T_{2}\Rightarrow T_{1}$. The converse implication does not hold however, and as a counter-example you may consider an infinite set with the co-finite topology, i.e. topology consisting of sets with finite complement plus the empty set. This space is not $T_{2}$ because any two open sets have an infinite intersection. However, it is $T_{1}$ because the complement of every singleton is open (thus you may take e.g. $U_{x}=\{y\}^{c}$ and $U_{y}=\{x\}^{c}$ for any $x\neq y$). Hence $T_{1}\not\Rightarrow T_{2}$.

Answer 2

$X$ is a $T_2$ (or Hausdorff) space if it has the following property:

whenever $x$ and $y$ are distinct points of $X$, there are open sets $U$ and $V$ in $X$ such that $x\in U$, $y\in V$, and $U\cap V=\varnothing$.

We say for short that $x$ and $y$ can be separated by disjoint open sets. For instance, the real line $\Bbb R$ with the usual topology is $T_2$. If $x,y\in\Bbb R$ with $x<y$, let $u$ be any real number such that $x<u<y$, let $U=(x-1,u)$ and let $V=(u,y+1)$; certainly $U$ and $V$ are open, and it’s easy to check that $x\in U$, $y\in V$, and $U\cap V=\varnothing$.

For a relatively simple example of a space that is not $T_2$, let $\tau$ be the cofinite topology on $\Bbb Z$: a subset $U$ of $\Bbb Z$ is open if and only if either $U=\varnothing$, or $\Bbb Z\setminus U$ is finite, i.e., $U$ contains all but finitely many integers. Suppose that $U$ and $V$ are open sets in this topology such that $0\in U$ and $1\in V$. Obviously $U$ and $V$ aren’t empty, so $\Bbb Z\setminus U$ and $\Bbb Z\setminus V$ are both finite sets; say $\Bbb Z\setminus U=\{n_1,\dots,n_r\}$ and $\Bbb Z\setminus V=\{m_1,\dots,m_s\}$. Pick any integer $k$ that is not in $\{n_1,\dots,n_r\}\cup\{m_1,\dots,m_s\}$; this set is finite, so there certainly is such an integer. But then $k\in U\cap V$, so $U\cap V\ne\varnothing$. Thus, there are no open sets $U$ and $V$ in this space such that $0\in U$, $1\in V$, and $U\cap V=\varnothing$. That shows that the space $\langle\Bbb Z,\tau\rangle$ is not $T_2$. (In fact you can check that if $U$ and $V$ are any non-empty open sets in this space, then $U\cap V\ne\varnothing$; essentially the same argument works.)

The space $\langle\Bbb Z,\tau\rangle$ is $T_1$, however. This simply means that

whenever $x$ and $y$ are distinct points of $X$, there is an open set $U$ in $X$ such that $x\in U$ and $y\notin U$.

You might say that $x$ can be separated from $y$ by the open set $U$. To see that this is the case with $\langle\Bbb Z,\tau\rangle$, let $n$ be any integer, and let $m$ be any other integer. Let $U=\Bbb Z\setminus\{m\}$; then by definition $U$ is open, and it’s clear that $n\in U$ and $m\notin U$.

A weaker property yet is the $T_0$ property. Superficially it looks a bit like the $T_1$ property: $X$ is $T_0$ if

whenever $x$ and $y$ are distinct points of $X$, either there is an open set $U$ such that $x\in U$ and $y\notin U$, or there is an open set $U$ such that $y\in U$ and $x\notin U$.

In other words, either $x$ can be separated from $y$ by and open set, or $y$ can be separated from $x$ by an open set, but not necessarily both. (In a $T_1$ space you can do both, though you may not be able to do it simultaneously with disjoint open sets.) The simplest example of a $T_0$-space that is not $T_1$ is the Sierpiński space, whose underlying set is $\{0,1\}$, and whose open sets are $\varnothing,\{1\}$, and $\{0,1\}$. The open set $\{1\}$ separates $1$ from $0$, but there is no open set separating $0$ from $1$: there is no open set that contains $0$ but not $1$. Thus, this space is $T_0$ but not $T_1$.

These are three of a rather large collection of separation axioms for topological spaces.