Conditions on a topological space implying that it has a minimal basis This is a question I've had in the back of my mind for a while, motivated by curiosity. 
Let $(X,\tau)$ be a topological space, and consider $\mathfrak{F}=\{B\subseteq\mathcal{P}(X):B \text{ is a basis for }\tau\}$ partially ordered by inclusion. Are there reasonable conditions on $\tau$ that ensure the existence of minimal elements in $(\mathfrak{F},\subseteq)$?  (I know that if $\tau$ is finite, then there is a minimal basis for $\tau$, but this is hardly interesting.) 
In particular, does $\Bbb R$ with the standard topology have a minimal basis?
 A: Here is a partial answer. We first give an answer for $\mathbb R$ and then a quick generalisation:
Lots of spaces do not satisfy this property. Take $X=\mathbb R$ and suppose such a minimal basis $\mathcal B$ exists. Pick $\varnothing\ne B\in\mathcal B$ and find $x<y$ inside it. Let $U=B\cap (-\infty, x+\frac23(y-x)),$ $V= B\cap (x+\frac13(x-y),\infty)$ be open. Then $B\ne U,V\ne\varnothing$ and $U\cup V=B$ and so by expressing $U,V$ as unions from the basis, we can express $b$ as the union of those unions. Hence $\mathcal B-B\subset\mathcal B$ is a basis but is strictly smaller.
For a general space $X$ the following properties are sufficient for $X$ to not admit a minimal basis:


*

*$X$ is a $T_{2{^1\!/_2}}$ space (points are separable by closed sets)

*$X$ is nonempty

*$X$ does not have the discrete topology.


The proof is the basically the same except you take just $x\ne y$ and use $U=B\cap(X-U'), V=B\cap(X-V')$ where $U',V'$ are a separation of $x,y$ by closed sets.
Properties 2 and 3 are equivalent to “any basis must have an element of cardinality at least 2”
If $X$ has the discrete topology then the set of singletons is a minimal basis.
I do not know about such minimal bases for other spaces.
A: I am unaware of any "nice" characterisation of spaces with minimal bases, but using the following one is able to show that a wide class of spaces do not have this property. It's not so much a characterisation of such spaces as a characterisation of the (necessarily unique) minimal base of a topological space which has one.
Given a topological space $X$, we call $x \in X$ a base point of $X$ if it  has a smallest open neighbourhood in $X$. In this case we denote this smallest open neighbourhood by $U(x)$.
Theorem. A topological space $X$ has a minimal base iff $\mathcal{M}(X) = \{ U(x) : x\text{ is a base point of }X \}$ is a (minimal) base for $X$.
Before we really begin, note the following observations.
Fact 1. If $\mathcal{B}$ is a minimal base for $X$, then for each $\mathcal{B}_0 \subseteq \mathcal{B}$, if $\bigcup \mathcal{B}_0 \in \mathcal{B}$, then $\bigcup \mathcal{B}_0 \in \mathcal{B}_0$.


*

*proof. Suppose that $\mathcal{B}_0 \subseteq \mathcal{B}$ is such that $U = \bigcup \mathcal{B}_0 \in \mathcal{B}$ and $U \notin \mathcal{B}_0$. Then as every set in the base $\mathcal{B}$ is the union of a subfamily of $\mathcal{B} \setminus \{ U \}$, it follows that $\mathcal{B}_0 \setminus \{ U \}$ is a base for $X$, contradicting the minimality of $\mathcal{B}$!


Fact 2. If $x$ is a base point for a topological space $X$, then $U(x) \in \mathcal{D}$ for every base $\mathcal{D}$ of $X$.


*

*proof. There must be a $V \in \mathcal{D}$ with $x \in V \subseteq U(x)$, but by definition of $U(x)$ it must be that $V = U(x)$.


We now turn to the proof of the theorem.


*

*(⇒) Suppose that $X$ has a minimal base $\mathcal{B}$. Given $U \in \mathcal{B}$, I claim that $U = U(x)$ for some (base point) $x \in X$. Otherwise for each $x \in U$ there is a $V_x \in \mathcal{B}$ such that $V_x \subsetneq U$. But then $\bigcup_{x \in U} V_x = U$, contradicting Fact 1!
Therefore $\mathcal{B} \subseteq \mathcal{M}(X)$, and since by Fact 2 we have $\mathcal{M}(X) \subseteq \mathcal{B}$, it follows that $\mathcal{B} = \mathcal{M}(X)$, as desired (i.e., $\mathcal{M}(X)$ is a minimal base for $X$).

*(⇐) Suppose that $\mathcal{M}(X)$ is a base for $X$. Using Fact 2 we have that $\mathcal{M}(X)$ is a subset of every base for $X$, and so $\mathcal{M}(X)$ is in fact a minimal base for $X$.

With regards to $\mathbb{R}$ under the usual topology, it is easy to show that it has no minimal base, since it has no base points. In fact, something much stronger can be said.
Proposition. If $X$ is a T1-space with a minimal base, then $X$ is discrete.


*

*proof. Suppose that $X$ is a T1-space. If $x \in X$ is non-isolated, then it can be shown that $x$ has no smallest open neighbourhood.  (Every open neighbourhood of $x$ is infinite, by and removing finitely many points distinct from $x$ we get a smaller open neighbourhood.)  If $x \in X$ is isolated, then $\{ x \}$ is the minimal open neighbourhood of $X$; i.e., $x$ is a base point of $X$, and $U(x) = \{ x \}$.
Then $\mathcal{M}(X) = \{ \{ x \} : x\text{ is isolated in }X \}$, and is a base for $X$ exactly when every point of $X$ is isolated; i.e., if $X$ is discrete.

One interesting class of spaces which have minimal bases is the class of Alexandrov spaces. A topological space $X$ is Alexandrov if it satisfies any/all of the following equivalent conditions (among many more):


*

*The family of open subsets of $X$ is closed under arbitrary (nonempty) intersections.

*The family of closed subsets of $X$ is closed under arbitrary unions.

*Every $x \in X$ has a smallest open neighbourhood, i.e., every $x \in X$ is a base point of $X$.


Using the last property above and the Theorem it is easy to show that every Alexandrov space has a minimal base. There are examples of spaces with a minimal base which are not Alexandrov.
Example. The family $\mathcal{B} = \{ [x,+\infty) : x \in \mathbb{Q} \}$ is a minimal base for a topology on $\mathbb{R}$. This topological space is not Alexandrov, as no irrational number is a base point.
