# Is a basis for a given topology always closed under finite intersections?

Define a basis $S$ for a given topology $\delta$ on $X$ as a subset of $\mathcal{P}(X)$ which satisfies the following conditions: $S \subseteq \delta$ and, for every $U \in \delta$ and every $p \in U$, there is a $V \in S$ such that $p \in V$ and $V \subseteq U$. It seems clear that, from this definition, it follows that every $U \in \delta$ will be equal to $\bigcup\limits_{i \leq n} V_i$ for some $V_1, \dots, V_n \in S$. My question is: does it also follow from this definition that, if $V_1, \dots, V_n \in S$, then $V_1 \cap \dots \cap V_n \in S$? Again, it's obvious that, for each $p \in V_1 \cap \dots \cap V_n$, there will be a $C \in S$ such that $p \in C$. But it doesn't seem clear to me that $S$ is necessarily closed under finite intersections. Am I missing something? Or is there a quick counterexample to this?

Clearly, the system of all open intervals of lengths $$1/n$$, $$n=1,2,\dots$$, is a base for the usual topology on $$\mathbb R$$. $$S=\{(a,a+1/n); a\in\mathbb R\}.$$ This base $$S$$ is not closed under finite intersections: $$(0,1/2)\cap(x,x+1/2)=(x,1/2)$$ for every $$x\in(0,1/2)$$. If we choose $$x$$ which is not of the form $$1/2-1/n$$, then this intersection does not belongs to $$S$$.
The usual topology on $\mathbb{R}$, for example, is generated by the basis $S$ consisting of all open intervals. But there are many open sets that are not a union of finitely many intervals. For example there are unbounded open sets.
The answer to the question in the title is also no. For example, choose a single element of $S$ above and remove it to obtain $S'$. $S'$ still generates the same topology (since the interval you threw away can be easily obtained as a union of two smaller intervals) but it is missing an intersection in $S'$ (since the interval you threw away can be easily obtained as an intersection of two larger intervals).
• About the first one: yes, you're correct, I messed up the indexes. What I meant is not that $U = \bigcup\limits_{i \leq n} V_i$, but that $U = \bigcup\limits_{i \in I} V_i$, where $I$ is some index (which may be an ordinal). But thanks a lot for the answer. There is probably some mistake in the exercise I was asked to do, then. May 1, 2014 at 5:45