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why $\mathbb{Q}$ and $\mathbb{Z}$ are zero-dimensional spaces ?

My attempt : Definition of zero-dimensional spaces:

A topological space $(X, \tau)$ is said to be zero-dimensional if there is a basis for the topology consisting of clopen sets

we know that in discrete space all basis for the topology consisting of clopen sets.so here i can said that obviously $\mathbb{Z}$ will be zero-dimensional spaces since $\mathbb{Z}$ induce discrete topology

But im confused about $\mathbb{Q}$ because it is neither closed nor open

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    $\begingroup$ $\Bbb Q$ and $\Bbb Z$ have a topology induced from the usual topology on $\Bbb R$. And $\Bbb Q$ is both closed and open in $\Bbb Q$. Try showing that the set of open intervals with irrational endpoints intersected with $\Bbb Q$ is a basis of clopen sets. $\endgroup$ – Elliot G Aug 14 '20 at 5:09
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    $\begingroup$ For any topological space $X$, the whole space is both closed and open, so that's not an issue. You need to find a basis for $\mathbb Q$ so that each basis element is both closed and open. Do you know any bases for $\mathbb Q$? Can they be adjusted? $\endgroup$ – Cheerful Parsnip Aug 14 '20 at 5:09
  • $\begingroup$ okks u mean $(\sqrt a,\sqrt b) \cap \mathbb{Q}$ @CheerfulParsnip $\endgroup$ – jasmine Aug 14 '20 at 5:13
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    $\begingroup$ There are more irrationals than just square roots... $\endgroup$ – Henno Brandsma Aug 14 '20 at 7:36
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The set $\mathcal{B}=\{(a,b)\mid a, b \in \Bbb P\}$ forms a base for the usual topology on $\Bbb R$ (where $\Bbb P = \Bbb R \setminus \Bbb Q$ is the set of irrationals).

For each $(a,b) \in \mathcal{B}$ it’s clear that $(a,b) \cap \Bbb Q = [a,b]\cap \Bbb Q$ and so the set $\{B \cap \Bbb Q\mid B \in \mathcal{B}\}$ is a base for the subspace topology of $\Bbb Q$ that consists of closed-and-open (clopen) sets.

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