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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?

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2 Answers 2

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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 $S$.

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Your first comment is false. Every open set will be a union of basis elements, but not necessarily finitely many basis elements.

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).

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  • $\begingroup$ 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. $\endgroup$
    – Nagase
    May 1, 2014 at 5:45

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