does every topology have a basis?

This might be a silly question, but i was wondering, is there any topology that cannot be generated by a basis? if not, given a topology, is there a reliable way of figuring out a basis for it? it probably matters if the set $X$ the topology is on is countable or not, right? Would it matter if the topology itself is countable? Thanks for any help/advice/feedback. :)

Sincerely,

Vien

• I wonder if you would like to consider the following: Let $\mathcal B$ be the family of all bases for a certain topology. Order $\mathcal B$ by inclusion. For some topologies there is a minimal element of $\mathcal B$, but that is not the situation one normally expects. You could look at circumstances under which $\mathcal B$ has minimal elements, or even a unique minimum element. – MJD Sep 10 '12 at 5:37

Added: On any set $X$ the indiscrete topology $\{X,\varnothing\}$ has only itself as base. There are other examples. For instance, for $n\in\Bbb Z^+$ let $V_n=\{1,\dots,n\}$, and let $\tau=\{\varnothing,\Bbb N\}\cup\{V_n:n\in\Bbb Z^+\}$; then $\tau$ is a topology on $\Bbb N$ whose only base is itself: none of the sets $V_n$ can be written as a union of the other sets in the topology, and $\Bbb N$ is the only open set containing $0$.
However, if each point of the space $\langle X,\tau\rangle$ has an open nbhd that is not the whole space, then $\tau\setminus\{X\}$ is always a base of $\tau$ different from $\tau$ itself. In particular this is true of every $T_1$ space with more than one point.
• @Brian M.Scott: I know it's been a while since you answered this, but I hope you can clarify something about your answer. I thought that the answer (to the original post) should be "yes", because given a topology $\tau$, the family $\tau \smallsetminus \{\emptyset \}$ is a basis and it's strictly coarser then $\tau$. So in both of your examples there exists a basis other then the topology itself. What am I missing? – Ludolila May 19 '14 at 12:42