Existence of non-Borel subset

Question

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!

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

2

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?

edited Apr 13 '17 at 12:21

Community

1

answered Nov 10 '14 at 22:49

PhoemueX

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Existence of non-Borel subset

1 Answers1