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I am reading "Topology without tears " book, and got confused about proposition 2.3.4 page 60:

$\tau_1 =\tau_2$ iif

1) for each $B \in \mathscr B_1 $ and $ \forall x \in B , \exists B' \in \mathscr B_2 $ such that $ x\in B' \subseteq B $

2) for each $B \in \mathscr B_2 $ and $ \forall x \in B , \exists B' \in \mathscr B_1 $ such that $ x\in B' \subseteq B $

I need intuition about this proposition! (If only part)

We are trying to prove two topologies are equivalent, so first we prove $\tau_1$ is subset of $\tau_2$ . Stop! Please explain what we are doing with basis of topologies here? Then book says by definition of basis and consideration of (1) above B is open in $\tau_2 $. Why?

Then book concludes $\mathscr B_1$ is $\subseteq \tau_2 $. Why?

Please expand your explanation part by part.

Thanx

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Suppose (1) is true. For each $x\in B$ let $B_x'\in\mathscr B_2$ such that $x\in B_x'\subseteq B$. So, $$B=\bigcup_{x\in B}B_x'$$.

Hence $B$ is open in $\tau_2$. Now if $U$ is open in $\tau_1$, it is a union of elements of $\mathscr B_1$ and hence a union of elements of $\mathscr B_2$. Therefore, $\tau_1\subseteq\tau_2$.

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Because if $\mathscr B$ is a basis for $\tau$, then every open set $U\in\tau$ is a union of elements in $\mathscr B$ if and only if $\mathscr B$ is a basis for $\tau$: $U=\bigcup_i B_i$, for $B_i\in\mathscr B$.

For part 1.) Let $U_1\in\tau_1$. Then $U_1=\bigcup_i B_i$ for $B_i\in\mathscr B_1$. But for each $B_j$, $B_j=\bigcup_{x\in B_j}B_x'$, where $B_x'\in\mathscr B_2$ and $x\in B_x'\subseteq B_j$. Since you can do this for each $B_j$, you can write $U_1$ as union of elements of $\mathscr B_2$, so $U_1\in\tau_2$. Since $U_1$ is arbitrary, $\tau_1\subseteq\tau_2$. Same thing for the other direction.

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  • $\begingroup$ I don't get $ B_j = \bigcup B'_x $ part. Where does this one comes? $\endgroup$ – Omid Nov 1 '15 at 1:18
  • $\begingroup$ Now l got it... so basically we are trying to show any elements of first topology can be constructed by basis of second topology? $\endgroup$ – Omid Nov 1 '15 at 1:21
  • $\begingroup$ Yes, that's the idea. Each $U$ of the first topology $\tau_1$ is constructed from elements of the first basis $\mathscr B_1$, but each element of the first basis is constructed from elements of the second basis $\mathscr B_2$, which means $U$ is constructed from elements of $\mathscr B_2$ and hence $U\in\tau_2$ $\endgroup$ – user153582 Nov 1 '15 at 1:39

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