Question about two sub bases generating a topology For the following questions, is it asking me to show the two subbgases generate the same topoloy.  If so, do I show that both subbases are both a base for the topology X and that each is a subset of the other.
Let $\mathcal{L_1}$ and $\mathcal{L_2}$ be collections of subsets of a set $X$ such that $X= \cup\{S\in\mathcal{L_1}\} = \cup\{S\in\mathcal{L_2}\}.$  Moreover, suppose that:
(1)  for each $S_1 \in\mathcal{L_1}$ and for each $x\in S_1$ there exists $S_2 \in\mathcal{L_2}$ such that $x\in S_2$ and $S_2 \subset S_1,$ and
(2)  for each $S_2 \in\mathcal{L_2}$ and for each $x\in S_2$ there exists $S_1 \in\mathcal{L_1}$ such that $x\in S_1$ and $S_1 \subset S_2.$
Then $\mathcal{L_1}$, $\mathcal{L_2}$ are subbases for the same topology on X.
The definition for a sub-basis is as follows:
Let $(X,\mathcal{F})$ be a topological space.  A subcollection $\mathcal{L}$ of $\mathcal{F}$ is a subbasis for $\mathcal{F}$ provided the family of all finite intersections of members of $\mathcal{F}$ is a basis for $\mathcal{F}.$
Thank you in advance
 A: Notation: $[A]^{<\omega}$ is the set of finite subsets of $A.$
For $i\in \{1,2\}$ let $T_i$ be the topology generated by the sub-base $L_i.$ Then $B_i=\{\cap S:S\in [L_i]^{<\omega}\}$ is a base for $T_i.$
If $i\in \{1,2\}$ and $x\in U_i\in T_i$ then there exists $b\in B_i$ with $x\in b\subset U_i,$ and  there exists $S\in  [L_i]^{<\omega}$ such that $\cap S=b.$  This requires $x\in \sigma\in L_i$ for each $\sigma \in S.$ 
Now the hypothesis implies that for each $\sigma  \in S$ we may take $\sigma'\in L_{3-i}$ with $x\in \sigma'\subset  \sigma.$
So let $U_{3-i}=\cap \{\sigma':\sigma\in S\}.$
We have $x\in U_{3-i}\subset \cap S=b\subset U_i.$ 
We have $\{\sigma':\sigma \in S\}\in [L_{3-i}]^{<\omega},$ so $U_{3-i}\in B_{3-i}\subset T_{3-i}.$ So we have $U_{3-i}\in T_{3-i}.$
So whenever $i\in {1,2}$ and $x\in U_i\in T_i$ there exists $U_{3-i}\in T_{3-i}$ such that $x\in U_{3-i}\subset U_i.$
With $i=1$ this implies $T_1\subset T_2$ and with $i=2$ this implies $T_2\subset T_1.$
A: Yes.
Hint: Specifically, you have to prove from the conditions that any finite intersection $\cap_{i\le n} S_i$ with $S_i\in\mathcal L_1$ is a union of sets of the form $\cap_{i\le k} T_i$ with $T_i\in\mathcal L_2$ and $k\in\Bbb N$. 
