Basis of topological space

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 $$\tilde{I}:=\{(a,b)\;|\;-\infty How do you verify that this family checks the properties $$(a)$$ and $$(b)$$? Thanks!

• What's your difficulty in that? – Berci Nov 3 '18 at 8:15
• @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 – Jack J. Nov 3 '18 at 8:21
• An example is not a proof. – William Elliot Nov 3 '18 at 8:24
• @William Elliot Exactly! – Jack J. Nov 3 '18 at 8:26

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 $$a:=\max(a_1,a_2), \ b:=\min(b_1,b_2)$$.
• For $\epsilon>0$ fixed $x\in (x-\epsilon,x+\epsilon)$. – Jack J. Nov 3 '18 at 8:45
• Yes, for instance $x\in (x-1,x+1)$ – Berci Nov 3 '18 at 8:53