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An arithmetic progression in $\mathbb{Z}$ is a set

$A_a,_b=\bigg\{\dots,a-2b,a-b,a,a+b,\dots\bigg\}$ with $a,b\in\mathbb{Z}$ and $b\neq0.$ prove that the collection of arithmetic progressions

$B=\bigg\{A_a,_b|a,b\in\mathbb{Z}\ and\ b\neq0\bigg\}$ is a basis for topology on $\mathbb{Z}$

any hint

Eklavya
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2 Answers2

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It is easy to note that $n\in A_{a, b}$ if and only if $n\equiv a$ $mod \ b$ and that solving two congruence equations will give you another congruence equation(if solvable)

chan kifung
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You need to show to things: first that $\mathbb{Z}$ is covered by members of $B$, which is easy (hint: $b=1$ is allowed).

Second that every intersection of two basic sets $A_{a,b}$ and $A_{c,d}$ is a union of members from $B$ as well. In this case, $A_{a,b} \cap A_{c,d}$ is either empty (which is OK), or is again of the same form, which you can try to show using elementary number theory.

Henno Brandsma
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  • in the second step i am getting stuck – Eklavya Feb 06 '16 at 09:19
  • What have you tried? In fact, it is enough to show that if $x$ is in the intersection, then there is some $A_{e,f}$ with $x \in A_{e,f} \subseteq A_{a,b} \cap A_{c,d}$, which is weaker and easier. – Henno Brandsma Feb 06 '16 at 09:20
  • i am geting stuck in getting structure of intersection of two basis elements in this case – Eklavya Feb 06 '16 at 09:23
  • Hint: if $x$ differs by a multiple of $b$ from $a$, and by a multiple of $d$ from $c$, what about the greatest common divisor of $b$ and $d$ ? – Henno Brandsma Feb 06 '16 at 09:26