2
$\begingroup$

I got the following statement:

For each set $N \subseteq \mathbb R^n$ with cardinality of the continuum $\#N=\mathfrak c$ there is a subset $M \subseteq N$ with $M \notin \mathcal B(\mathbb R^n)$.

Can somebody tell me why this is true? I thought of different cardinalities, but that doesn't seem to work out.. Thanks in advance!

$\endgroup$
2
$\begingroup$

The cardinality of the Borel sigma algebra is the cardinality of the continuum (see Cardinality of Borel sigma algebra).

The cardinality of the set of all subsets of $N$ is the cardinality of the power set of the continuum.

Hence?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.