Definitions. Suppose $X$ is a topological space.
- $w(X)=\min\{|\mathcal B|:\mathcal B$ is a base for $X\}+\omega$
- $e(X)=\sup\{|D|:D\subseteq X$ is closed and discrete$\}+\omega$
- $K(X)$ is the collection of all compact subsets of $X$
- $\mathbb R ^*=\beta\mathbb R \setminus \mathbb R$
The collection of all open subsets of $\mathbb R$ has cardinality $\mathfrak c$. The canonical basis for $\beta \mathbb R$ consists of sets of the form
$$\{p\in\beta\mathbb R :(\exists L\in p)(L\subseteq U)\}$$ where $U$ is open in $\mathbb R$. Thus $w(\beta\mathbb R) \leq \mathfrak c$ ( I suspect equal).
Now I appeal to a theorem, which may or may not be a deep result (I haven't read the proof) in cardinal invariants.
Theorem. If $X$ is compact $T_1$ then $|K(X)|\leq 2^{e(X)\cdot w(X)}$.
Note that $e(\mathbb R ^*)=\omega$ (really finite, but in the definition we require that it be at least $\omega$): If $D$ is an infinite discrete closed subspace of $\mathbb R^*$, then $D$ contains a copy of $\beta\omega$, contradicting $D$ discrete.
Combining everything we have $|K(\mathbb R ^*)|\leq 2^{\omega\cdot \mathfrak c}=2^\mathfrak c$ (whereas $|\mathcal P (\mathbb R^*)|=2^{2^\mathfrak c}$).
Questions:
1) Is my calculation correct?
2) Is there an easier way to get my final conclusion? I feel there must be, but maybe not. We know $\beta\mathbb R$ has a countable dense subset, but I didn't see how to use that.
Edit: I might be an idiot. Why does the following not work: $\beta\mathbb R$ has a countable dense subset $\mathbb Q$, and we can uniquely identify an open subset of $\beta \mathbb R$ by its intersection with $\mathbb Q$. Thus there are only $2^\omega$ open subsets of $\beta\mathbb R$, whence there are also only $2^\omega$ closed subsets of $\beta\mathbb R$... clearly false.