Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let's say we have some topological space.

Axiom $T_1$ states that for any two points $y \neq x$, there is an open neighborhood $U_y$ of $y$ such that $x \notin U_y$.

Then we say that a topological space is $T_4$ if it is $T_1$ and also satisfies that for any two closed, non-intersecting sets $A,B$, there are open neighborhoods $U_A,U_B$ respectively, such that $U_A\cap U_B = \emptyset$.

Could anyone give an example of a topological space which satisfies the second condition of $T_4$, but which is not $T_1$?

share|cite|improve this question
It's sometimes helpful to know not every author includes T1 as a requirement for T4. This is one area where the conventions vary from book to book. For example, Counterexamples in Topology does not require a T4 space to be T1. – Carl Mummert Oct 2 '10 at 10:57
up vote 20 down vote accepted

Consider a topological space $X$ with two (different) points and the trivial topology (that is, only $\emptyset$ and $X$ are open sets).

This space satisfies the second condition of $T_4$: the only two closed non-intersecting sets are $\emptyset$ and $X$ and you can take as open neighborhoods $U_\emptyset = \emptyset$ and $U_X = X$.

But $X$ is not $T_1$, as you can easily check.

$T_1$ (also called Fréchet space) is equivalent to the fact that every point is a closed set (exercise: prove this!). So, if you had a regular space ($T_3$) $X$ that was not $T_1$, you could not say that $X$ is $T_2$ (or Hausdorff); that is, you couldn't say that $T_3 \Rightarrow T_2$.

So, it is necessary to add the $T_1$ condition to $T_3$ and $T_4$ in order to have the beautiful sequence of implications that makes all of us happy :-)

$$ T_4 \ \text{(normal)} \ \Longrightarrow \ T_3 \ \text{(regular)} \ \Longrightarrow \ T_2 \ \text{(Hausdorff)} \ \Longrightarrow \ T_1 \ \text{(Fréchet)} $$

share|cite|improve this answer

As an addition: the separating closed disjoint sets part is often called normality (a space is normal if it satisfies this), so $T_4$ is normal plus $T_1$, and similarly for regular: $T_3$ is regular and $T_1$. Spaces that have no disjoint non-empty closed sets (besides the trivial topology, we have examples like $\mathbf{N}$ with the topology generated by the sets of the form $U(n) = \{ k : k \ge n \}$ e.g.) trivially satisfy normality. $T_4$ is to avoid these pathologies: the extra $T_1$ ensures that at least all finite sets are closed, so we have some "relevant" closed sets to apply normality to...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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