A topological space with cardinality strictly less than its weight?

My book states that

if $X$ is compact, then $w(X) \leq |X|$.

This leads me to wonder, is there a nice example of a (non-compact) topological space where $w(x) > |X|$ holds?

The weight of a topological space is the smallest cardinality of a basis of the topology.

The maximal weight of a space $X$ is $2^{|X|}$ and in this answer I give some examples (with proofs) of spaces that realise this maximum. E.g. a countable dense subset of $\{0,1\}^{\mathbb{R}}$ in the product topology has weight $|\mathbb{R}|$.

The bound $w(X) \le |X|$ for compact spaces follows from the equality $nw(X) =w(X)$ for (locally) compact spaces (and it also holds for metrisable spaces and orderable spaces), so many standard examples don’t work.

I believe the club-generated topology on $\omega_1$ is an example (a set is open in this topology iff it is a union of sets which are unbounded and closed in the order topology - this does indeed form a topology, since the intersection of two clubs is a club). Given any $\omega_1$-sequence of open sets $C_\eta$, I can build a club (hence open) set which doesn't contain any of them, as follows:

• We begin at stage $0$ with $D_0=\emptyset$.

• At stage $\alpha+1$, pick some $\beta\in C_\alpha$ greater than $\sup(D_\alpha)$, and let $D_{\alpha+1}=D_\alpha\cup\{\beta\}$ (so since we keep adding bigger ordinals, we'll keep $\beta$ out of the club we're building). This is possible since $C_\alpha$, being a union of clubs, is unbounded.

• At stage $\lambda$ limit, let $D_\lambda=(\bigcup_{\alpha<\lambda} D_\alpha)\cup\{\sup(\bigcup_{\alpha<\lambda} D_\alpha)\}$.

Let $D=\bigcup_{\eta<\omega_1} D_\eta$. $D$ is club, hence open, and doesn't contain any of the $C_\eta$s. So $\{C_\eta:\eta<\omega_1\}$ was not a base for the club-generated topology on $\omega_1$.

It's worth noting that this argument does use a bit of choice: in the successor steps of our construction, we're assuming that there are in fact ordinals bigger than $\sup(D_\alpha)$. This uses the fact that $\omega_1$ is not a countable union of countable sets, which surprisingly is not provable in ZF alone. Interestingly, ZF does prove that $\omega_2$ is not a countable union of countable sets. So that's nice.

EDIT: As Asaf points out, if $\omega_1$ has countable cofinality then "club" doesn't really make sense (every singleton is now the intersection of two clubs, so the topology is discrete).