Could anyone please check my proof of the second assertion in the title and prove the first one? (It will be enough to work with the set of all real numbers). Let the topological space be denoted as $T = (R^1, \tau)$ where $\tau$ consists of the "set of all subsets of $R^1$."
My proof of the second part:
i) Since every single-element set is considered open in this topological space and since a base is a collection of open sets where every open set in $T$ is a union of some sets of this collection, any base must contain all the single-element sets (all real numbers). But this is naturally uncountable.
ii) For the first part, we would need to prove that given any $x \in R^1$ there is a countable family of open sets such that for every open set $G$ that contains $x$ there is a set $O$ in this family that is contained in $G$. We know that any set $G$ is already an open set in $T$. But if the statement is true wouldn't it imply that the set of all sets in $R^1$ that contain an arbitrary pont $x$ is countable? Is this corect?
Happy Good Friday 🙂
Best Answer
In fact $\{\{x\}\}$ is a (finite, hence countable) local base at $x$.
If $\mathcal{B}$ is any base for $\tau$, then for all $x$ there must be a $B \in \mathcal{B} $ such that $x \in B \subseteq \{x\}$, which implies that $B =\{x\} \in \mathcal{B}$, so that the base has size at least $|\Bbb R|$. So an uncountable set in the discrete topology doesn’t have a countable base.