Generally it is understood that a $T_1$ space is a space in which for two points $x,y$ there exist open sets $U,V$ such that $x \in U, y \notin U, y \in V, x \notin V$. A $T_2$ or Hausdorff space is a space where in addition $U \cap V = \varnothing$.

But for $T_3$ there appear to be at least two seemingly different definitions!


A space is $T_3$ iff it is $T_1$ and regular where regular is used to mean that every neighborhood contains a closed neighborhood.


A space is $T_3$ iff it is $T_0$ and regular where regular is used to mean that given any nonempty closed set $F$ and any point $x$ not in $F$, there exists a neighbourhood $U$ of $x$ and a neighbourhood $V$ of $F$ that are disjoint.

It is not obvious to me that these two defintions are equivalent. It is also not clear to me if the two definitions or regular space are equivalent.

Please could someone explain these definitions to me? I'm a bit confused.


2 Answers 2


These are equivalent. First of all, regularity can be formulated as "every point has a neighborhood basis of closed sets" or as "every point and every closed set not containing that point can be separated by open neightborhoods." Secondly, $T_0\!+$regular implies $T_1\!+$regular implies $T_0\!+$regular.

  • $\begingroup$ Great clear answer, thanks. Meanwhile unfortunately I found a third definition here according to which a space is $T_3$ iff it's $T_2$ and regular. Is this also equivalent to the other two definitions? $\endgroup$
    – newb
    Feb 8, 2014 at 18:44
  • $\begingroup$ Apparently, yes. It says so in the paragraph. $\endgroup$
    – newb
    Feb 8, 2014 at 18:45
  • 1
    $\begingroup$ Yes they are equivalent because in a $T_1$ space singletons are closed, and by regularity the points can be separated by open neighborhoods. Thus it is Hausdorff. $\endgroup$
    – user123641
    Feb 8, 2014 at 21:21

Note that if $X$ is T0 and "regular", given distinct $x,y \in X$ there is an open set $U$ containing one but not the other. Without loss of generality, let's assume that $x \in U$. We will now show that there is an open $W$ containing $y$ but not $x$ (this will show that $X$ is T1). Note that by the regularity condition there is an open $V$ with $x \in V \subseteq \overline{V} \subseteq U$, and so $X \setminus \overline{V}$ is an open set containing $y$ but not $x$.

Similarly, we can show that a space which is T0 and "completely regular" is T3 1/2.

However, this is false for T4: the 2-point Sierpiński space, $X$, is T0 but not T1. It is, however, "normal": if $F,E$ are disjoint closed subsets of $X$ then at least one is empty, and so the disjoint open sets $\varnothing$, $X$ separate $F$ and $E$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.