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Theorem. Let $(X, \mathcal{G})$ a topological space.

$(i)$ Let $\mathcal{B}\subseteq\mathcal{G}$ a basis of $(X,\mathcal{G})$. Then,

$(a)$ $\mathcal{B}$ is a coverage of $X$;

$(b)$ for each $B_1,B_2\in\mathcal{B}$, $B_1\cap B_2\ne\emptyset$ and for each $x\in B_1\cap B_2$ exists $B\in\mathcal{B}$ such that $x\in B\subseteq B_1\cap B_2$.

$(ii)$ Let $\mathcal{B}\subseteq \mathcal{P}(X)$ a family of nonempty set for which they are valid $(a)$ and $(b)$, then exists a unique topology $\mathcal{G}$ on $X$ of which $\mathcal{B}$ is basis.

Question. Let the family set $$ \tilde{I}:=\{(a,b)\;|\;-\infty<a<b<+\infty\}\subseteq\mathcal{P}(\mathbb{R}) $$ How do you verify that this family checks the properties $(a)$ and $(b)$? Thanks!

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  • $\begingroup$ What's your difficulty in that? $\endgroup$ – Berci Nov 3 '18 at 8:15
  • $\begingroup$ @BerciMy difficulty is to strictly prove that those properties are true. With examples I can understand that they are true, but I do not know if it should be prove $\endgroup$ – Jack J. Nov 3 '18 at 8:21
  • $\begingroup$ An example is not a proof. $\endgroup$ – William Elliot Nov 3 '18 at 8:24
  • $\begingroup$ @William Elliot Exactly! $\endgroup$ – Jack J. Nov 3 '18 at 8:26
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To prove (a), assume we are given an $x\in\Bbb R$. Can you write up a basis element that contains $x$?
To prove (b), assume $x$ is in both intervals $(a_1,b_1)$ and $(a_2,b_2)$. Then consider the following positive numbers: $$L:=\min(x-a_1,\,x-a_2),\ \ \ R:=\min(b_1-x,\,b_2-x)$$ Or, take directly $a:=\max(a_1,a_2), \ b:=\min(b_1,b_2)$.

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  • $\begingroup$ For $\epsilon>0$ fixed $x\in (x-\epsilon,x+\epsilon)$. $\endgroup$ – Jack J. Nov 3 '18 at 8:45
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    $\begingroup$ Yes, for instance $x\in (x-1,x+1)$ $\endgroup$ – Berci Nov 3 '18 at 8:53
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    $\begingroup$ In this case the base is even closed under finite intersections (when the intersection is non-empty). $\endgroup$ – Henno Brandsma Nov 3 '18 at 9:20

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