# Power set of Cantor set has cardinality greater than that of reals

I am reading the example of "Boreal measure is not complete" from Wiki:

https://en.wikipedia.org/wiki/Complete_measure

In the first example, it says

The power set of the Cantor set has cardinality strictly greater than that of the reals. Thus there is a subset of the Cantor set that is not contained in the Borel sets.

My questions are the following:

1. Why "Cantor set has cardinality strictly greater than that of the reals"?
2. Why "there is a subset of the Cantor set that is not contained in the Borel sets"? Can anyone give me an example of this?

Thanks!

• The Cantor set has cardinality equal to the cardinality of the reals, so its power set has greater cardinality. The cardinality of the collection of Borel sets is equal to the cardinality of the continuum. – user296602 Dec 13 '18 at 2:35
• @T.Bongers Just sincerely ask a dumb question: "The Cantor set has cardinality equal to the cardinality of the reals" since both are equal to infinity? – sleeve chen Dec 13 '18 at 2:37
• Since both are equal to continuum. Infinities don't all have the same size. – Mason Dec 13 '18 at 2:39
• No - after all, that would trivialize the rest of your problem, because it would mean all infinite sets have equal cardinality. The Cantor set can be identified with the ternary numbers in $[0, 1]$ missing a $1$ in their base 3 expansion, which is naturally equinumerous with all of $[0, 1]$ by base 2 expansion. – user296602 Dec 13 '18 at 2:44

As mentioned in the comments, the Cantor set has cardinality $$2^{\aleph_0}$$ (i.e. the same cardinality as the reals). Thus its power set has cardinality $$2^{2^{\aleph_0}},$$ which by Cantor's theorem is greater than the cardinality of the reals. On the other hand, it can be shown that there are only $$2^{\aleph_0}$$-many Borel sets (this isn't trivial). Thus, there are more subsets of the Cantor set than there are Borel sets, so there is a subset of the Cantor set that is not a Borel set.