# Difference between $T_0$ spaces and $T_1$ spaces

Can anybody tell me the difference between $T_0$ space and $T_1$ space?

I can see both have the same definition. But I know a space is $T_1$ if and only if every singletons are closed.

If $X=\{a,b,c\}$,what are the what are the Topologies on $X$ that are $T_0$ but not $T_1$?

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Generally, things with the same definition have either the same name or very different names. Things which are named $T_0$ and $T_1$ in the same field (topology) are very unlikely to have the same definition. It should be clear that over a century of mathematicians would have noticed that already. –  Asaf Karagila Nov 5 '12 at 23:17

They don't have the same definition, in fact. A space $X$ is $T_0$ if for all distinct $x,y\in X$, at least one of them has a neighborhood that doesn't contain the other. A space $X$ is $T_1$ if for all distinct $x,y\in X$, each of them has a neighborhood that doesn't contain the other.

For examples of topologies on $\{a,b,c\}$ that are $T_0$ but not $T_1$, consider $5$ through $8$, here.

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No, they do not have the same definition, though you may have been given the definitions in forms that are superficially very similar. One common definition of the $T_1$ property is that $X$ is $T_1$ if for each pair of distinct points $x$ and $y$ there is an open set $U$ such that $x\in U$ and $y\notin U$. What you may not realize right away is that because this applies to every pair of points, it applies to $y$ and $x$ as well, so that there is also an open set $V$ such that $y\in V$ and $x\notin V$. In other words, given two points in a $T_1$-space, each has an open neighborhood that does not contain the other. This is equivalent to saying that all singletons in $X$ are closed, but this does require a proof.

A space $X$ is $T_0$ if for each pair of distinct points $x$ and $y$ in $X$ at least one of $x$ and $y$ has an open neighborhood that does not contain the other; it doesn’t have to work both ways. Thus, the topology

$$\Big\{\varnothing,\{a\},\{a,b\},\{a,b,c\}\Big\}\tag{1}$$

is a $T_0$ topology on $\{a,b,c\}$: given any two points in the space, the one that comes earlier in the alphabet has an open neighborhood that does not contain the other. It is not a $T_1$-space, however, because (for instance) $c$ has no open neighborhood that does not contain $a$: the only open neighborhood of $c$ is $\{a,b,c\}$. Thus, when we consider the pair of points $c$ and $a$ (in that order), there is no open $U$ such that $c\in U$ and $a\notin U$. And you can also see that the only closed singleton in this space is $\{c\}$: the sets $\{a\}$ and $\{b\}$ are not closed in this topology.

The only $T_1$ topology on $\{a,b,c\}$ is the discrete topology, in which every subset of $\{a,b,c\}$ is open.

Added: It’s only mildly painful to find all of the $T_0$-topologies on $X=\{a,b,c\}$. One is the discrete topology, but it’s also $T_1$ (and more). There are altogether $6$ $T_0$ topologies homeomorphic to $(1)$, one for each permutation of $X$. There are $6$ more $T_0$ topologies homeomorphic to

$$\Big\{\varnothing,\{a\},\{a,b\},\{c\},X\Big\}\;;$$

there are $3$ ways to choose the non-isolated point, and then $2$ ways to choose the isolated point with which to pair it. And that’s it: one $T_1$ topology, and $12$ topologies that are $T_0$ but not $T_1$, $6$ in each of two homeomorphism classes. All other topologies on $X$ fail to be even $T_0$.

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@ Brian,this is a wonderful answer Brian,thank you very much! –  ccc Nov 6 '12 at 0:46
@ccc: You’re very welcome! –  Brian M. Scott Nov 6 '12 at 0:48

In $T_0$ space we just want that whenever we are given two distinct points there is some open set in which one of the points lie and the other don't. In a $T_1$ space, however, we also require that all singletons are closed, so we can actually tell what is this open set (the complement of the singleton of one of the points).

For example $\{0,1\}$ with the topology $\{\varnothing,\{0,1\},\{1\}\}$ is $T_0$ but not $T_1$ because whenever we are given two distinct points we have to have that one of them is $1$ and the other is $0$, therefore $\{1\}$ is open as wanted. This topology is not $T_1$ because $\{1\}$ is not closed, though.

As for your given $X$ there are only finitely many topologies, but for $T_1$ topologies you have quite the constraints (all singletons are closed) whereas for $T_0$ you have slightly more options but still not too many.

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