$B = \{\{x\}|x\in X\}$ Prove B is a basis for a topology on $X$

general-topology

Say $X$ is a set and we know $\mathcal{B} = \{\{x\}|x\in X\}$.

I'm trying to show that:

(a) $\mathcal{B}$ is a basis for a topology on $X$

(b) that $\mathcal{B}$ generates the discrete topology.

The criteria for a basis are:

For each $x\in X$ there is some $B\in \mathcal{B}$ such that $x\in B$

For any $B_1$,$B_2 \in \mathcal{B}$ and any $x \in B_1 \cap B_2$, there is some $B \in \mathcal{B}$ such that $x \in B⊆B_1\cap B_2$.

For part (a), I'm honestly somewhat lost. $\mathcal{B}$ is the set of points, so there must be subsets that have fewer points than $\mathcal{B}$, but how do I know if the second criterion for a basis applies without knowing the size of a subset?

For part (b), I know that the discrete topology is the largest topology containing all subsets as open sets. Assuming that I know part (a) has given me the basis for a topology:

We have a collection of all one-point subsets on $X$, which is the basis for the discrete topology by definition.

Proving if $X$ is discrete then every set of one element is open is easy. If $X$ is discrete, we know this by definition.

I'm less sure how to prove $X$ is discrete because all of the one-point subsets are open. I would guess that you would take some arbitrary set in $X$, something like $U$, and say that $U = \bigcup_{u \in U}\{u\}$. Then you know it's a union of open sets. Because it was arbitrary, $X$ only contains open sets and must be discrete.

Best Answer

For (a) just note that for any $x \in X$ we have $x \in \{x\} \in \mathcal{B}$. That's all for the first. The second criterium is trivial, as $x \in B_1 \cap B_2$ immediately implies that $B_1 = B_2 = \{x\}$ (as both sets only have one element, and it must be $x$) so if that happens we just take $B_3 = \{x\}$ too.

For (b): the topology generated by $\mathcal{B}$ is by definition the set of all unions of members of $\mathcal{B}$ and any $A \subseteq X$ is such a union: $A = \bigcup \{\{x\}\mid x \in A\}$. So the generated topology is just the power set of $X$, i.e. the discrete topology.

This whole exercise is a set-theory triviality.