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Please check my proof

Define arithmetic progressions

$A={A_{a,b}|a,b\in Z,b\neq 0}$

$=$

$....a-2b,a-b,a,a+b,...$

Let all of union of arithmetic progression $A_{a,b}=Z$

Let $x\in Z$ then $x\in (a\pm bn)$ and $(a\pm bn)\in A_{a,b}$

Suppose

$A_{1}={A_{i,j}|i,j\in Z,j\neq 0}$ $=$$....i-2j,i-j,i,i+j,...$

$A_{2}={A_{c,d}|c,d\in Z,d\neq 0}$ $=$ $....c-2d,c-d,c,c+d,...$

and

$A_{1},A_{2}\subset A$

it's exist

$A_{3}={A_{e,f}|e,f\in Z, f\neq 0}$

such that

$x\in A_{3}$ and $x\in A_{3}\subset A_{2}\cap A_{1}$

for some $A_{1}=A_{2}$

therefore A is basis for topology on $Z$

monkeylover401
  • 55

1 Answers1

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So the proposed base is all sets $A(a,b)$ where $a,b \in \mathbb{Z}, b \neq 0$, and where $$A(a,b) = \{a + nb: n \in \mathbb{Z}\}$$

is an arithmetic progression with start point $a$ and "jump" $b$.

We have to check the usual two properties:

a. Every $a \in \mathbb{Z}$ is in some base element: this is clear because $\mathbb{Z}= A(0,1)$, so one base element suffices.

b. For any two elements $B_1= A(a_1,b_1)$ and $B_2 = A(a_2,b_2)$ and any $x \in B_1 \cap B_2$ we must find some $B_3$ containing $x$ sitting inside this intersection. We know that $x = a_1 + n_1b_1 = a_2 + n_2b_2$ for some $n_1, n_2 \in \mathbb{Z}$. But then $x \in A(x, b_1b_2) \subseteq B_1 \cap B_2$, as e.g. for any $n \in \mathbb{Z}$: $x + nb_1b_2 = a_1 + n_1b_1 + nb_1b_2 = a_1+ (n_1 + b_2n)b_1 \in A(a_1,b_1)$ and likewise for $A(a_2,b_2)$.

So the arithmetic progressions define a unique topology on $\mathbb{Z}$.

Henno Brandsma
  • 242,131