1979
DOI: 10.1080/00029890.1979.11994929
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Measure and Cardinality

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Cited by 4 publications
(3 citation statements)
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“…Consequently, validating Proposition 1 translates to showing that any given w i does not lie in the span of W span −i ; that is, w i and W span −i are linearly independent. First, it is easy to note that P (w 1 = 0) = 1, since the Lebesgue measure of a singleton set is zero [79]; thus, P (w 1 / ∈ W span −i ) = 1. Furthermore, for p ∈ {2, ..., n − 1}, let W −1,p = {w 1 , ..., w p } be the set of first p vectors, excluding vector i, such that W −1,p is linearly independent with a probability of 1.…”
Section: Main Findings Onmentioning
confidence: 99%
“…Consequently, validating Proposition 1 translates to showing that any given w i does not lie in the span of W span −i ; that is, w i and W span −i are linearly independent. First, it is easy to note that P (w 1 = 0) = 1, since the Lebesgue measure of a singleton set is zero [79]; thus, P (w 1 / ∈ W span −i ) = 1. Furthermore, for p ∈ {2, ..., n − 1}, let W −1,p = {w 1 , ..., w p } be the set of first p vectors, excluding vector i, such that W −1,p is linearly independent with a probability of 1.…”
Section: Main Findings Onmentioning
confidence: 99%
“…Consequently, proofing Proposition 1 translates to showing that any given w i does not lie in the span of W span −i ; that is, w i and W span −i are linearly independent. First, it is easy to note that P (w 1 = 0) = 1, since the Lebesgue measure of a singleton set is zero [57]; thus, P (w 1 / ∈ W span −i ) = 1. Furthermore, for p ∈ {2, ..., n−1}, let W −1,p = [w 1 , ..., w p ] be the matrix whose columns are the first p vectors, excluding vector w i , such that the columns of W −1,p are linearly independent with a probability of 1.…”
Section: Appendixmentioning
confidence: 99%
“…It is still true without the continuum hypothesis, but it is more difficult to prove. See[1].3 In general, still assuming axiom of choice, you cannot get a set with some infinite cardinality κ, by taking finite products of sets with smaller cardinality, or by taking union of µ < κ sets with smaller cardinality.…”
mentioning
confidence: 99%