# Is there a way to describe all possible topologies on $\mathbb{R}$ with $T_0$ axiom?

Let's consider a topological space $(\mathbb{R}, \tau)$ that meets $T_0$ axiom. What can we say about $\tau$? I have a wild guess that there are not so many of them. I failed to construct topology weaker than the standard topology. I can think of only four ways to define $\tau$:

1. $\tau_1 = \tau(\{(a,b)\colon\ a,b\in\mathbb{R}\})$ - standard topology

2. $\tau_2 = \tau(\{[a,b)\colon\ a,b\in\mathbb{R}\})$

3. $\tau_3 = \tau(\{(a,b]\colon\ a,b\in\mathbb{R}\})$

4. $\tau_4 = 2^\mathbb{R}$ - discrete topology

Is there anything else?

Edit: okey, I just thought of $\tau_5 = \tau_1\cup\tau_2$ and $\tau_5 = \tau_1\cup\tau_3$

Edit 2: okey, these were not really new topologies...

• The cofinite topology, the co-countable topology, the co-measure-0 topology, the co-bounded topology. There are huge huge huge numbers of topologies that satisfy $T_0$. (All of the above contain the co-finite sets.) You are only thinking geometrically, but the set $\mathbb R$ is just a set when talking about topologies on it - it has no other structure. – Thomas Andrews Feb 17 '16 at 23:39
• The number of topologies depends only on the cardinality, not in the particular set. So there are lots and lots of them. – Matt Samuel Feb 17 '16 at 23:39
• Yes, for example, since $\mathbb R^n$ is the same cardinality as $\mathbb R$, there is a topology on $\mathbb R$ that makes the space homeomorphic to $\mathbb R^n$. – Thomas Andrews Feb 17 '16 at 23:40
• Now, if you want topologies such that the binary operations $+,\times :\mathbb R\times\mathbb R\to\mathbb R$ are continuous, you might get fewer topologies. – Thomas Andrews Feb 17 '16 at 23:43
• By the way, $\tau_1\subset \tau_2$ and $\tau_1\subset \tau_3$, so you "Edit" comment isn't actually giving new topologies. This is because $(a,b)=\bigcup_{n} [a+\frac{1}{n},b)$. – Thomas Andrews Feb 17 '16 at 23:51

Any topology in which all finite sets are closed is $T_0$, and all of your topologies are such topologies.

The co-finite topology is the smallest such topology - the open sets are the sets $U$ with either $U=\emptyset$ or where $\mathbb R\setminus U$ is finite.

One class of topologies that does not contain the co-finite sets are the "order" topologies. The most common (and actually, useful) one is the left- and right-open topology:

$$\tau_{-}=\{(-\infty,a)\mid a\in\mathbb R\cup\{\pm\infty\}\}\\ \tau_{+}=\{(a,+\infty)\mid a\in\mathbb R\cup\{\pm\infty\}\}\\$$

These are actually useful - they are related to left- and right- continuity.

Give a topology $\tau$ on $X$, and an element $x\in\mathbb X$, there is a notion of localizing $\tau$ to $x$. An open set in $\tau_x$ is either any set not containing $x$ or any set which contains a $U\in \tau$ with $x\in U$.

Functions $X\to Y$ are "continuous at $x$" if $(X,\tau_x)\to Y$ is continuous. We can show easily that $\tau_x$ is a $T_0$ space even when $\tau$ was not.

Then it turns out that $\tau = \bigcap_{x\in X} \tau_x = \tau$. This roughly means that continuity on the entire space is the same as continuity at every point.

Now, an interesting question might be: What are the topologies ($T_0$ or otherwise) on $\mathbb R$ such that the standard binary operations $+,\times:\mathbb R\times\mathbb R\to\mathbb R$ are continuous. I don't have an answer to that question.

There are many, many more than that. OEIS 000798 says there are already $355$ topologies on four labeled points and OEIS 006057 says there are $137$ of them on four labeled points that satisfy $T_0$ through $T_4$ For $\Bbb R$ the number is clearly uncountably infinite and I would guess it is $2^{\mathfrak c}$.

• It’s even bigger: $2^{2^\mathfrak{c}}$. – Brian M. Scott Feb 18 '16 at 4:50

There are $2^{2^{\mathfrak{c}}}$ such topologies (this is also an upper bound as a topology on $\mathbb{R}$ is a subset of the power set of $\mathbb{R}$). Indeed, choose a free ultrafilter $U$ on $\mathbb{R}$ and set $\tau_U = U\cup \{\varnothing\}$. Then $\tau_U$ is a $T_0$-topology. By Pospíšil's theorem, there are $2^{2^{\mathfrak{c}}}$ such ultrafilters.