Does the set of open sets in $\mathbb{R}$ is countable?

Question

I want to prove that the cardinality of Borel $\sigma $ algebra is $\aleph$, using the next proposition:
If $ E \subset P(X) $ is infinite, and the cardinality of E is $\aleph $ , the $\sigma$ algebra generated by E is of cardinality $\aleph$.
Im pretty stuck with the step of proving that the generators of that $\sigma$ algebra, which are the open sets of Borel, is of the requested cardinality. Here, I'm taking the standard topology on R, where $\aleph$ is the cardinality of continoum.

$\aleph$ does not refer to a single cardinal. Do you mean $\aleph_0$ instead? Also, open under what topology? The usual one? — ViHdzP, Nov 16 '20 at 18:34
@URL: $\aleph$ is sometimes used to denote the cardinality of the continuum. This goes back to Hasudorff. — Asaf Karagila, Nov 16 '20 at 18:36
When you write $\aleph$, I think you mean $\aleph_0$, but the are are uncountably many open sets in $\mathbb{R}$ with the usual topology. All the sets $(0,x)$ where $x>0$ are open, so there are at least $\mathfrak{c}$ open sets. — saulspatz, Nov 16 '20 at 18:36
@saulspatz: \mathfrak c. Also my comment. The Borel algebra is not countable, just to clarify. — Asaf Karagila, Nov 16 '20 at 18:36
Ron, I think you want $E\subseteq\mathcal P(X)$, not $E\in\mathcal P(X)$. — Asaf Karagila, Nov 16 '20 at 18:38

score 1 · Accepted Answer · answered Nov 16 '20 at 18:48

1

An open subset of $\mathbb{R}$ is a countable union of open intervals, so the open intervals generate the Borel sets. There are $\aleph$ open intervals, so the theorem you quote completes the proof.

answered Nov 16 '20 at 18:48

saulspatz

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Does the set of open sets in $\mathbb{R}$ is countable?

1 Answers1