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The collection $\mathcal{B} = \{ \{x\} : x \in X \}$ is a basis for the discrete topology on a set X. If X is a finite set with n elements, then clearly $\mathcal{B}$ also has n elements. Is there a basis with fewer than n elements that generates the discrete topology on X?

What I think: No, $T_{discrete} = P(X)$ which includes all possible subsets of X including the sets of singletons $\{ \{x\} : x \in X \}$ so any bases of $T_{discrete}$ must have at least n elements. Am I in the right direction ?

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  • $\begingroup$ You should be more explicit in justifying why a basis of the discrete topology must contain the singletons. $\endgroup$ – angryavian May 25 '16 at 0:48
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If $\mathcal{B}'$ is a basis, then in particular every element of $\mathcal{B}$ is a union of elements of $\mathcal{B}'$. But a singleton cannot be a union of proper subsets, so $\mathcal{B} \subset \mathcal{B}'$ and $\mathcal{B}'$ has at least $n$ elements.

As an alternative proof, we could observe that the number of possible unions that we can form from a collection of $k$ subsets is at most $2^k$. Therefore, if a collection of $k$ sets forms a basis, we must have $2^k \geq 2^n$, so $k\geq n$.

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