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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.

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Is there an easier way to get my final conclusion?

Let $o(X)$ denote the cardinality of the topology of the space $X$ (i.e., it is the number of open subsets of $X$.) The same notation is used in Juhasz' book Cardinal Functions in Topology - Ten Years Later.

Then we have inequality $o(X)\le 2^{w(X)}$, which follows from the fact that every open set can be obtained as a union of some system of basic sets.

If we use $w(\beta\mathbb R)\le\mathfrak c$, then we get $$o(\beta\mathbb R)\le 2^{\mathfrak c}.$$

So there are at most $2^{\mathfrak c}$ open subsets in $\beta\mathbb R$. Clearly, the number of closed subsets is the same. And since we are working with subsets of a compact space, compact subsets are precisely closed subsets, so we get $|K(\beta\mathbb R)|\le 2^{\mathfrak c}$.

Since weight of a subspace is less or equal to the weight of the whole space, we also have $w(\mathbb R^*)\le w(\beta\mathbb R) \le \mathfrak c$ and we can use the same argument to get $$|K(\mathbb R^*)|=o(\mathbb R^*) \le 2^{\mathfrak c}.$$


On the other hand, we have $o(X)\ge |X|$ for any $T_1$ space. Since $|\beta\mathbb R|=2^{\mathfrak c}$ we get $$o(\beta\mathbb R)=2^{\mathfrak c}.$$ For the proof of cardinality of $\beta\mathbb R$ see: Stone–Čech compactification of $\mathbb{N}, \mathbb{Q}$ and $\mathbb{R}$

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  • $\begingroup$ Thanks, I thought of that before sleeping last night, but I will give credit for the answer anyway! $\endgroup$ – Forever Mozart Jun 30 '14 at 14:50

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