Prove that there exists a unique topology $T$ on $X$ so that $P$ is its subbasis and that
$ card (T) = 2^n$. Let $X$ be a non-empty set and $P$ a partition of $X$ so that $ card (P) = n$, for $ n \in \mathbb{N}$. Prove that there exists a unique topology $T$ on $X$ so that $P$ is its subbasis and that
$ card (T) = 2^n$. 
It's easy to prove that there exists a unique topology $T$ on $X$ so that $P$ is its subbasis, because trivially the union of all elements in $P$ covers the whole $X$. As for the second part I'm a bit confused, because the basis for said topology consists of every finite intersection of the elements of $P$, but as all of those elements are disjunct sets, we are left with only trivial intersections? So the basis would be $P$? How can then the cardinality of the topology $T$ be $2^n$?  This would have to mean that $T$ is discrete. 
Is there a mistake in the question or am I thinking wrong?
 A: The set of finite intersections of members of $P$ is just $P$ again (as we only have the one-element trivial intersections, while taking at least two different members only leaves us with $\emptyset$. So the corresponding base is also $P$ and then one just has to realise that if $P=\{P_1,\ldots,P_n\}$ all unions  of subfamilies from $P$ is just indexed by subsets of $\{1,\ldots n\}$: we get open sets $O(I) = \bigcup_{i \in I} P_i$ for all subsets $I$ of $\{1,\ldots,n\}$, of which there are exactly $2^n$ many. (So $(\emptyset)=\emptyset$ and $O(I)=X$, $O(\{i\})=P_i$ etc.) And each of them is different: if $i \in I\setminus J$ then $x \in O(I)\setminus O(J)$ for any $x \in P_i$, e.g. So $I \in \mathscr{P}(I) \to O(I)$ is a bijection between all open sets and the powerset of an $n$-point set. All these unions together form the topology $\mathcal{T}$ induced by $P$ and it thus has $2^n$ members.
This topology is only discrete when $P$ consists of only singletons sets. If $P_i$ is a non-singleton member of $P$ and $x\neq y$ are in $P_i$, then $\{x\}$ is not open in $\mathcal{T}$ (not a union of partition elements, the only one that contains $x$ is $P_i$ and then $y$ would also have been in the union) but is open in the discrete topology on $X$.
A: The topology $T$ is formed by all the possible unions of the sets forming the partition, and all the sets $t \in T$ can be written uniquely as unions of $p \in P$. So there is a bijection between $T$ and $\mathcal{P}(P)$:
$$t = \cup_{i \in I}p_i \rightarrow \{ p_i | i \in I\}$$
In general, the topology $T$ is not discrete: a topology $T_d$ on $X$ is discrete when $T_d = \mathcal{P}(X)$, and $T_d$ has cardinality $2^{\#X}$, not $2^n$!
