I recently started studying topology (Royden and Fitzpatrick's Real Analysis – 4th edition) and I stumbled upon this exercise:
"Show that the trivial (indiscrete) topology on a set $\, X \,$ has a unique base."
Consider a topological space $(X, \mathcal{T})$, where $\mathcal{T} = \{\, \varnothing, \,X \, \}$.
My first reaction was "The unique base obviously is $\mathcal{B}_1 = \{X \}$". But the more a thought about it, the more it became apparent that $\mathcal{B}_2 = \{\, \varnothing, \,X \, \}$ could also be a base. So, now, I'm wondering if it is something definitional that I'm missing. The way the book defines a base is as follows:
Definition: For a topological space $(X, \mathcal{T})$ and a point $x \in X$, a collection of neighborhoods of $x$, $\mathcal{B}_x$ is called a base for the topology at $x$ if for any neighborhood $A$ of $x$, there is $B \in \mathcal{B}_x$ such that $B \subseteq A$. A collection of open sets $\mathcal{B}$ is called a base for the topology $\mathcal{T}$ provided it contains a base for the topology at each point $x$.
Note that considering $\mathcal{B}_1$ or $\mathcal{B}_2$, which are both collections of open sets of $\mathcal{T}$, for any arbitrary $x \in X$ we can define a collection of neighborhoods of $x$, $\mathcal{B}_x = \{ X \}$. Then considering any neighborhood $A$ of $x$ (In this case the only possible neighborhood is $A = X$, we can take $X \in \mathcal{B}_x$ and it is clear that $X \subseteq X$. By definition this would imply that there is not a unique base for $\mathcal{T}$, in fact both $\mathcal{B}_1$ and $\mathcal{B}_2$ are basis for this topology.
If anyone can point out the error I made in my explanation, or (preferably) side with my answer, it would be greatly appreciated.
Best Answer
One definition of a basis is a set $\mathcal{B}$ is a subset of $\mathcal{T}$ such that every member of $\mathcal{T}$ is a union of members of $\mathcal{B}$. This is equivalent to your definition if you allow the empty union. The only thing I can think is that this author does not, but that contradicts their own definition. If you don't allow the empty union, then clearly the empty set is mandatory.