# Compare the vertical intervals topology with the standard topology on $\mathbb{R}^2$.

1- Show that the collection $$C=\lbrace (-\infty,q) \subset \mathbb{R} | q\in \mathbb{Q} \rbrace$$ is a basis for the topology $$\tau = \lbrace (-\infty,p) \subset \mathbb{R} | p\ in \mathbb{Q} \rbrace$$.

Let $$A \in \tau$$, $$\forall x\in A, \exists w=(-\infty, q)$$ s.t. $$x\in (-\infty, q)$$ from the dense of $$Q$$.

2-a) Show that the collection $$B= \lbrace \lbrace a \rbrace × (b,c) \subset \mathbb{R}^2 | a,b,c \in \mathbb{R} \rbrace$$ of the vertical intervals in the plane is a basis for a topology on $$\mathbb{R}^2$$.

Let $$w_1, w_2 \in B$$ such that $$w_1=\lbrace a \rbrace × (b,c), w_2= \lbrace x \rbrace × (y,z)$$

It should to be that $$a=x$$, to obtain that $$w_1 \cap w_2\neq \phi$$

$$w_1 \cap w_2 = \lbrace a \rbrace × (y,c)$$ if $$b

Or $$w_1 \cap w_2 = \lbrace a \rbrace × (b,z)$$ if $$y

So, $$\forall w_1, w_2 \in B, \forall m\in w_1 \cap w_2, \exists w_3 \in B$$ s.t. $$x\in w_3 \subset w_1 \cap w_2$$

2-b) Compare the vertical intervals topology with the standard topology on $$\mathbb{R}^2$$.

I think, for all $$A=\lbrace a \rbrace × (b,c) \in B$$, take the open disk D is centered on $$(a,\frac{c-b}{2})$$ and have the radius $$\frac{c-b}{2}+1$$, such that $$A\in D$$, as a result, “the vertical topology” $$\subset$$ “the standard topology”.

• You might add the proof-verification tag to your question as that seems to be what you're asking for – postmortes Feb 4 at 7:52
• @postmortes sorry, I don’t understand you, since I don’t speak English well. – Dima Feb 4 at 10:57

The fact that $$B$$ is a basis for a topology is pretty clear. Now you have to prove that any open subset of $$\mathbb R^2$$ for the standard topology can be written as a union of elements of $$B$$.
Take an open disk $$D$$ centered on $$(c_1,c_2)$$ with radius $$r>0$$. You’ll verify that
$$D = \bigcup_{c_1-r\le x \le c_1+r} L_x \cap D$$
where $$L_x$$ is the vertical line passing through the point $$(x,0)$$ and that $$L_x \cap D$$ is an element of $$B$$.
• Thanks, is this to prove that $B$ is a basis ? – Dima Feb 4 at 11:48
• @Dima This is to prove that $B$ generates the standard basis of $\mathbb R^2$. – mathcounterexamples.net Feb 4 at 16:12