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I was reading a set theory book and they claim that there are as many open sets in $\mathbb R$ as real numbers (usual topology).

I tried using the base of intervals with irrational extremes but found nothing.

Any hint is welcome.

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    $\begingroup$ Every open subset of $\mathbb{R}$ can be written as a union of open intervals with rational endpoints. $\endgroup$ – saulspatz Mar 26 at 2:17
  • $\begingroup$ That's not true. $\mathbb{R}$ and $\mathbb{R} -\sqrt{2}$ are different open sets but contain the same set of rationals. $\endgroup$ – David Lui Mar 26 at 3:06
  • $\begingroup$ @DavidLui Thanks. :-( $\endgroup$ – user759562 Mar 26 at 3:13
  • $\begingroup$ @DavidLui They do indeed contain the same set of rationals, but that doesn't stop them from being expressed as unions of open intervals with rational endpoints. For example, $\mathbb R \backslash \{\sqrt{2}\}$ is the union of the open intervals $(-n, a_n)$ and $(b_n, n)$ where $a_n$ and $b_n$ are increasing and decreasing sequences, respectively, of rationals with limit $\sqrt{2}$. $\endgroup$ – Robert Israel Mar 26 at 3:49
  • $\begingroup$ Sine the other person deleted their post, I was responding to another comment saying that an open set was determined by the rationals that it contained. $\endgroup$ – David Lui Mar 26 at 4:03
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As |{ ($\infty$,r ) : r in R }| = c = |R|,
c <= |topology R|.

Each open set is a countable, pairwise disjoint,
union of open intervals and rays.

How many open intervals and rays are there?
How many finte collections of open intervals and rays and are there?
How many denumberable collections of open rays and are there?
c$^{|N|}$ = c.
Add them all together and what do you get?

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