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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 $b<y<c<z$

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

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”.

Is that true, pleased? Thanks.

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

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  • $\begingroup$ Thanks, is this to prove that $B$ is a basis ? $\endgroup$ – Dima Feb 4 at 11:48
  • $\begingroup$ @Dima This is to prove that $B$ generates the standard basis of $\mathbb R^2$. $\endgroup$ – mathcounterexamples.net Feb 4 at 16:12
  • $\begingroup$ Now I understand, Thank you so much. $\endgroup$ – Dima Feb 4 at 16:13

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