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Nov 18, 2019 at 18:00 history tweeted twitter.com/StackStats/status/1196488370924539906
Nov 18, 2019 at 17:27 comment added whuber In set theory, the cardinal number $2^n$ is defined to be the cardinality of the set of functions from a set of $n$ elements to a set of $2$ elements. Writing the latter set as $\{0,1\}$ (with no loss of generality) shows these functions can be identified with the indicator functions on $\mathcal B$--but an indicator function is determined by, and determines, the subset of $\mathcal B$ on which it equals $1,$ QED. In this way the quoted statement makes sense even when $\mathcal B$ is countably infinite.
Nov 18, 2019 at 16:12 answer added BCLC timeline score: 3
Nov 18, 2019 at 12:47 history became hot network question
Nov 18, 2019 at 8:21 answer added knrumsey timeline score: 9
Nov 18, 2019 at 6:15 vote accept Nemo
Nov 18, 2019 at 5:38 history edited gunes
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Nov 18, 2019 at 5:37 answer added gunes timeline score: 16
Nov 18, 2019 at 4:35 history asked Nemo CC BY-SA 4.0