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let X be a set, and $\mathfrak S$ denote a collection of subsets of X. If $\mathfrak S$ is a subbase for $\tau_1 $ and a base for $\tau_2$ show that $\tau_1=\tau_2$

My approach: suppose $H\in \tau_1$. since $\mathfrak S$ is a subbase of $\tau_1$, $H=\cup_i(S_{i_1}\cap S_{i_2}...\cap S_{i_{n_i}}) $ which $S_{i_k}\in \mathfrak S $.

and since $\mathfrak S $ is a base for $\tau_2$ and so $\mathfrak S\in\tau_2 $ for this reason each $S_{i_k}\in\tau_2$.therefore $H\in\tau_2$ and $\tau_1\subseteq\tau_2$

otherwise : suppose $H\in\tau_2$.since $\mathfrak S$ is a base of $\tau_2$, $H=\cup_i S_i$ where $S_i\in \mathfrak S $. we can write each $S_i$ as $S_i\cap S_i$ so H would equal to $\cup_i(S_i\cap S_i)$. so we get finite intersections. thus, we can write H as the union of finite intersections. so $H\in \tau_1$ and $\tau_2\subseteq\tau_1$.

is this correct?

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If $\mathfrak S$ is a base for $\tau_2$, it is also a subbase for $\tau_2$, since each union of elements of $\mathfrak S$ is in particular a union of finite intersections. Thus, $\mathfrak S$ is a subbase for both $\tau_1$ and $\tau_2$, so $\tau_1=\tau_2$.

Your proof is correct as well. You don't need to write $S=S\cap S$ though, since $S = \bigcap_{i=1}^1 S$ is trivially a finite intersection.

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