Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 2 / 13G (Further Analysis)

Define $\mathcal{T}$ to be the family of all unions of members of $\mathcal{B}$, including the empty union. We will verify that this is a topology and that $\mathcal{B}$ is a basis of a topology. The two conditions play different roles: the covering condition supplies the whole space, while the local intersection condition supplies closure under finite intersections.

By the covering condition, $X={\bigcup }_{B\in \mathcal{B}}B$ belongs to $\mathcal{T}$. The empty set belongs to $\mathcal{T}$ as the empty union. An arbitrary union of sets that are themselves unions of members of $\mathcal{B}$ is still a union of members of $\mathcal{B}$. Thus $\mathcal{T}$ contains $X,\varnothing$ and is closed under arbitrary unions, as required for a topology.

Take $U,V\in \mathcal{T}$ and $x\in U\cap V$. Their representations as unions provide $B_1,B_2\in \mathcal{B}$ with $x\in B_1\subseteq U$ and $x\in B_2\subseteq V$. The intersection condition supplies $B\in \mathcal{B}$ with $x\in B\subseteq B_1\cap B_2\subseteq U\cap V$.

It follows that $U\cap V$ is the union of all members of $\mathcal{B}$ contained in it, and hence belongs to $\mathcal{T}$. Induction gives closure under every finite intersection. Together with part (i), this proves that $\mathcal{T}$ is a topology. Each $B\in \mathcal{B}$ belongs to $\mathcal{T}$ as a union with one member, and every open set is a union of such members by construction. Thus $\mathcal{B}$ is a basis for $\mathcal{T}$. Here the printed subset sign is interpreted as inclusion, allowing equality, as in the usual basis criterion.

The lexicographic order is a total order. Each point $p=(a,b)$ lies in the order interval between $(a,b-1)$ and $(a,b+1)$, so the order intervals cover ${\mathbb{R}}^2$.

If $p$ lies in both $⟨x_1,y_1⟩$ and $⟨x_2,y_2⟩$, let $x=\max (x_1,x_2)$ and $y=\min (y_1,y_2)$ in that total order. Then $x<p<y$ andIn particular an order interval contains $p$ and is contained in the intersection. Both conditions of part (a) hold, proving that these intervals form a basis of a topology. The resulting topology is the lexicographic order topology on the real plane.

For every $a\in \mathbb{R}$, the nonempty order intervalis open in the lexicographic order topology on the real plane. The sets $U_a$ are pairwise disjoint.

Suppose $\{B_n:n\in \mathbb{N}\}$ were a countable basis of a topology. For each $a$, the basis of a topology property gives a member containing $(a,0)$ and contained in $U_a$. Let $n(a)$ be the least index of such a member. The same nonempty $B_n$ cannot lie in two disjoint $U_a$, so $a\mapsto n(a)$ is an injection from the uncountable set $\mathbb{R}$ into $\mathbb{N}$, a contradiction. This topology is not second countable.