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Let the plane $\mathbb R \times \mathbb R$, and endow it with the topology generated by the set of complement of 1 lines ( $y= ax+b; a,b\in\mathbb R$ ) . This set is a sub-basis, the question is, found a minimal basis $\mathcal B$, in the sense that given any basis $\mathcal A$ , contained in $\mathcal B$, then $\mathcal A=\mathcal B$.

I conjecture that the basis, is the basis generated by the sub-basis, I´ll like to see other answers to this question

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I can't make any sense out of this question. – Alexei Averchenko Aug 17 '11 at 9:08
@Alexei: I take Daniel to be asking the following question. Let $\mathscr{S}$ be the set of complements of straight lines in the plane, and take $\mathscr{S}$ as the subbase for a topology $\mathscr{T}$. Let $\mathscr{B}$ be the set of finite intersections of members of $\mathscr{S}$. If $\mathscr{B_1}$ is a minimal base for $\mathscr{T}$ with respect to inclusion, is it true that $\mathscr{B_1}=\mathscr{B}$? – Brian M. Scott Aug 17 '11 at 9:31
that is the question, find a minimal basis – Daniel Aug 17 '11 at 10:12
So what did you try, Daniel? – Alexei Averchenko Aug 17 '11 at 10:43

It’s useful first to determine exactly what the topology in question is. Let $\mathscr{L}$ be the set of lines in the plane, let $\mathscr{G} = \{\bigcup\mathscr{F}:\mathscr{F}\subseteq \mathscr{L}\text{ is finite}\}$, and let $\mathscr{K} = \{\bigcap\mathscr{C}:\mathscr{C}\subseteq\mathscr{G}\}$. Then $\mathscr{L}$ is a subbase for the closed sets of that topology, $\mathscr{G}$ is the associated base for the closed sets, and $\mathscr{K}$ is the family of closed sets itself.

Let $K = \bigcap\mathscr{C}$, where $\mathscr{C} \subseteq \mathscr{G}$. For each $C \in \mathscr{C}$ there is a finite $\mathscr{L}_C \subseteq \mathscr{L}$ such that $C = \bigcup\mathscr{L}_C$. Then $K = \{x \in \mathbb{R}^2:\forall C \in \mathscr{C} \exists L_x \in \mathscr{L}_C (x \in L_x)\}$. Thus, if $\Phi$ is the set of functions $\varphi:\mathscr{C} \to \bigcup\limits_{C \in \mathscr{C}} \mathscr{L}_C$ such that $\varphi(C) \in \mathscr{L}_C$ for every $C \in \mathscr{C}$, $K = \bigcup\limits_{\varphi\in\Phi}\bigcap\limits_{C\in\mathscr{C}}\varphi(C)$. Note that each of the sets $\bigcap\limits_{C\in\mathscr{C}}\varphi(C)$ is an intersection of lines and is therefore empty, a singleton point, or a line. Now fix $C_0 \in \mathscr{C}$. For any $\varphi \in \Phi$, $\bigcap\limits_{C\in\mathscr{C}}\varphi(C) \subseteq \varphi(C_0)$, i.e., $\bigcap\limits_{C\in\mathscr{C}}\varphi(C)$ is either empty, the line $\varphi(C_0)$, or the singleton of a point on that line. Let $G = \bigcup\{L \in \mathscr{L}_{C_0}:L \subseteq K\} \in \mathscr{G}$, and let $P = K\setminus G$.

Suppose that $P$ is infinite. $\mathscr{L}_{C_0}$ is finite, so $P \cap L_0$ is infinite for some $L_0 \in \mathscr{L}_{C_0}$. $L_0 \nsubseteq K$, so there is some $C_1 \in \mathscr{C}$ such that $L_0 \notin \mathscr{L}_{C_1}$. Similarly, $P \cap L_0$ is an infinite subset of $C_1$, so $P \cap L_0 \cap L_1$ is infinite for some $L_1 \in \mathscr{L}_{C_1}$. But this is absurd: the distinct lines $L_0$ and $L_1$ can intersect in at most one point. Thus, $P$ must be finite. It follows that every $K \in \mathscr{K}$ can be written in the form $K = G \cup P$, where $G \in \mathscr{G}^* \triangleq \mathscr{G} \cup \{\varnothing\}$ and $P$ is a finite subset of the plane.

Clearly each singleton belongs to $\mathscr{K}$, since it can be written as the intersection of two lines. Since $\mathscr{K}$ is closed under taking finite unions, it follows that $G \cup P \in \mathscr{K}$ for each $G \in \mathscr{G}^*$ and finite $P \subseteq \mathbb{R}^2$ and hence that the closed sets are precisely those of the form $G \cup P$ for $G \in \mathscr{G}^*$ and finite $P \subseteq \mathbb{R}^2$. The open sets are of course the complements of these, but it’s easier to continue to work with the closed sets.

$\mathscr{G}$ is a base for the closed sets, and the first question is whether it’s a minimal base. If not, we’d like to know whether there is a minimal base, and if so, what it is. It’s not hard to see that $\mathscr{G}$ isn’t minimal. Let $\mathscr{G}_0 = \mathscr{G} \setminus \mathscr{L}$; certainly $\mathscr{G}_0$ is closed under taking finite unions. For any $L \in \mathscr{L}$ pick distinct $L_1,L_2 \in \mathscr{L} \setminus \{L\}$; then $L \cup L_1$ and $L \cup L_2$ both belong to $\mathscr{G}_0$, and their intersection is $L$. Clearly, then, $\{\bigcap \mathscr{C}:\mathscr{C} \subseteq \mathscr{G}_0\} = \mathscr{K}$, and $\mathscr{G}_0$ is a base for the closed sets properly contained in $\mathscr{G}$. It takes a little more work to see that $\mathscr{K}$ has no minimal base.

Let $\mathscr{B}$ be a base for the closed sets; that is, $\mathscr{B} \subseteq \mathscr{K}$, $\mathscr{B}$ is closed under taking finite unions, and $\{\bigcap \mathscr{C}:\mathscr{C} \subseteq \mathscr{B}\} = \mathscr{K}$. Each $K \in \mathscr{K}$ can be written in the form $G_K \cup P_K$, where $G_K \in \mathscr{G}^*$, and $P$ is a finite subset of $\mathbb{R}^2$. For each $B \in \mathscr{B}$ let $$\mathscr{B}(B) = \left\{K \in \mathscr{B} \setminus \{B\}:G_K \subseteq B_K \land P_K \subseteq P_B\right\},$$ and say that $B \in \mathscr{B}$ is inaccessible from below if $B \ne \bigcup\mathscr{B}(B)$. Note that if $\mathscr{B}(B) = \varnothing$, then $B$ is inaccessible from below, and that if $B$ is inaccessible from below, $\mathscr{B}\setminus \{B\}$ is closed under taking finite unions.

Let $B \in \mathscr{B}$ be arbitrary, and let $x,y \in \mathbb{R}^2 \setminus B$. There are $\mathscr{C}_x,\mathscr{C}_y \subseteq \mathscr{B}$ such that $\bigcap \mathscr{C}_x = \{x\}$ and $\bigcap \mathscr{C}_y = \{y\}$. Let $\mathscr{C} = \{B \cup C: C \in \mathscr{C}_x \cup \mathscr{C}_y\}$; then $\mathscr{C} \subseteq \mathscr{B} \setminus \{B\}$, and $\bigcap \mathscr{C} = B$. It follows that if $B \in \mathscr{B}$ is inaccessible from below, $\mathscr{B}\setminus \{B\}$ is a base for $\mathscr{K}$ properly contained in $\mathscr{B}$. To complete the proof that $\mathscr{B}$ is not minimal, it suffices to show that some $B \in \mathscr{B}$ is inaccessible from below.

Fix $B_0 \in \mathscr{B}$ arbitrarily. If $B_0$ is inaccessible from below, we’re done, so assume that $B_0 = \bigcup \mathscr{B}(B_0)$. Then $\mathscr{B}(B_0)$ is finite and non-empty, so there is a $B_1 \in \mathscr{B}(B_0)$ that is minimal with respect to $\subseteq$. But then $\mathscr{B}(B_1) = \varnothing$, so $B_1$ is inaccessible from below, and we’re done.

Added: This actually shows that the space has no minimal base generated by a subbase, i.e., no minimal open base closed under taking finite intersections and no minimal closed base closed under taking finite unions. I have no idea why I restricted my attention to these bases. I’m leaving the argument, because I think that the result is at least mildly interesting in its own right, but it wasn’t necessary to work quite so hard: the observation that for each $B \in \mathscr{B}$ there is some $\mathscr{C} \subseteq \mathscr{B} \setminus \{B\}$ such that $B = \bigcap\mathscr{C}$ is enough to show that any single element can be removed from $\mathscr{B}$.

Even that is excessively specific. Let $\langle X,\mathscr{T} \rangle$ be a space; it’s not hard to see that $\mathscr{T}$ has a minimal base iff each $x \in X$ has a minimal open nbhd. If the space is $T_1$, as in the present example, this is the case iff the topology is discrete; $\omega$ with the initial segment topology is an example of a non-discrete $T_0$ space with a minimal base.

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I think there is no minimal basis. The basis you suggest is that of complements of graphs made with a finite number of straight lines.

The thing is you can dispose of the one-line graph complements from the basis, because you can generate those considering the union of three two-line graph complements (it's tricky, try it!).

Once you discover this trick, you can do the same to dipose of the two-line graph complements using three-line graph complements and so on.

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Your suspicion is correct, but your argument would at most show that no subset of the given base is a minimal base; it doesn’t deal with other possible bases. – Brian M. Scott Aug 18 '11 at 5:44

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