Cardinality of Elements in a Basis Call a topological space $X$ "large" if it satisfies the following propriety: $$\text{For all bases $B$ of $X$, }\exists U\in B \text{ such that } |U|=|X|$$
This condition is equivalent to saying that $\exists x\in X$ such that, if $x\in U$ for some open set $U$, then $|U|=|X|$ (I'm quite confident of equivalence, yet not sure). What other conditions are equivalent or sufficient for $X$ to be large? How can I tell if $X$ is large without having to look through each possible basis or each element?
 A: The equivalence you hypothesise I think is correct:
Suppose $$\exists x \in X: \forall O \in \mathcal{T}_X: x \in O \implies |X|  =|O|$$ 
holds, and $\mathcal{B}$ is any base. Then some base member contains $x$ and for this one we are done. This is a trivial direction.
If the condition does not hold, then any $x \in X$ has a "small" open neighbourhood $N_x$ with $|N_x| < |X|$. The set $\mathcal{B} = \{O \in \mathcal{T}_X: |O| < |X|\}$ is a base for $X$:  let $x \in O$, $O$ open. Then $N_x \cap O$ is "small" as $|N_x \cap O| \le |N_x| < |X|$ and sits between $x$ and $O$. This $\mathcal{B}$ refutes the requirement for the base.
So these are indeed equivalent.
I don't see any other essentially different one. 
If I may ask: why define this notion? What's your interest? Euclidean spaces obey it (but these are strongly homogeneous). 
In fact in any normed vector space $X$, every point has arbitrarily small neighbourhoods that are homeomorphic to $X$.
$\omega_1$ as an ordered space (the first uncountable ordinal) is locally countable, so does not obey it. Neither do discrete spaces of size 2 or more, as the singleton base shows.
