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Let $X = \mathbb Z$ (the integers), and for each $n \in \mathbb Z$, define $B_n = \{m \in \mathbb Z: m \geq n\}$.

Let $\mathscr B = \{B_n : n \in Z\}$.

Prove $\mathscr B$ is a base for a topology on $\mathbb Z$.

I understand to do this I need two things.

  1. For each $x\in X$ there is some $B\in\mathscr B$ such that $x\in B$, and

  2. For any $B_0,B_1\in\mathscr B$ and any $x\in B_0\cap B_1$, there is some $B\in\mathscr B$ such that $x\in B\subset B_0\cap B_1$.

I'm so confused where to start.

I have seen

Prove that B is a basis for a topology

but don't understand the answer.

Thanks

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k9b
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    To expand on Cameron’s answer just a little, you should look at $B_m\cap B_n$ for some specific choices of $m$ and $n$. What are $B_3\cap B_5$, $B_4\cap B_1$, and $B_7\cap B_2$? Can you find a general expression for $B_m\cap B_n$ in terms of $m$ and $n$? – Brian M. Scott Oct 08 '14 at 02:22

1 Answers1

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Well, start by taking some $x\in X.$ Can you find some $B_n$ such that $x\in B_n$? It may help you to consider a few specific examples of $x$ first. Put less formally, every element of $X$ needs to be an element of some $\mathscr B$-set.

As for the other part, the notation is probably confusing you. Instead, based on your post at present, you want to show that if $B_m,B_n\in\mathscr B$ and $x\in B_m\cap B_n$, then there is some $B_k\in\mathscr B$ such that $x\in B_k$ and $B_k\subseteq B_m\cap B_n$. Again, it may help to look at some specific examples, first. Put less formally, if any two $\mathscr B$-sets have a common element, then there needs to be some $\mathscr B$-set contained in those two $\mathscr B$-sets, and containing that common element.

P.S.: I'm fairly sure that you intended $X$ to be the set of all integers, rather than the real numbers, though the result still holds in any case.

Brian M. Scott
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Cameron Buie
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