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David Feldman
David Feldman
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For simplicity let us drop the condition that $a_\{n+1\}>a_n$$a_{n+1}>a_n$. It just means that we exclude a countable set in every step, that should not be very interesting. We can still consider the supremum of a sequence with respect to the well-order and ask wetherwhether it is a given countable ordinal $c$. This event has probability zero, since it is contained in the set of all sequences where the first value is contained in a certain countable set. This set is contained the countable union of sets of the form $\left\{a\right\}\times [0,1]^{\mathbb{N}}$, which have measure zero. QED. You simply constructed many different sets of probability zero containing sequences with infinitely many different values (but you enforce that the values are contained in a certain countable set, and there are many countable subsets of the reals). I do not think that this is very astonishing.

For simplicity let us drop the condition that $a_\{n+1\}>a_n$. It just means that we exclude a countable set in every step, that should not be very interesting. We can still consider the supremum of a sequence with respect to the well-order and ask wether it is a given countable ordinal $c$. This event has probability zero, since it is contained in the set of all sequences where the first value is contained in a certain countable set. This set is contained the countable union of sets of the form $\left\{a\right\}\times [0,1]^{\mathbb{N}}$, which have measure zero. QED. You simply constructed many different sets of probability zero containing sequences with infinitely many different values (but you enforce that the values are contained in a certain countable set, and there are many countable subsets of the reals). I do not think that this is very astonishing.

For simplicity let us drop the condition that $a_{n+1}>a_n$. It just means that we exclude a countable set in every step, that should not be very interesting. We can still consider the supremum of a sequence with respect to the well-order and ask whether it is a given countable ordinal $c$. This event has probability zero, since it is contained in the set of all sequences where the first value is contained in a certain countable set. This set is contained the countable union of sets of the form $\left\{a\right\}\times [0,1]^{\mathbb{N}}$, which have measure zero. QED. You simply constructed many different sets of probability zero containing sequences with infinitely many different values (but you enforce that the values are contained in a certain countable set, and there are many countable subsets of the reals). I do not think that this is very astonishing.

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The User
The User
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For simplicity let us drop the condition that $a_\{n+1\}>a_n$. It just means that we exclude a countable set in every step, that should not be very interesting. We can still consider the supremum of a sequence with respect to the well-order and ask wether it is a given countable ordinal $c$. This event has probability zero, since it is contained in the set of all sequences where the first value is contained in a certain countable set. This set is contained the countable union of sets of the form $\left\{a\right\}\times [0,1]^{\mathbb{N}}$, which have measure zero. QED. You simply constructed many different sets of probability zero containing sequences with infinitely many different values (but you enforce that the values are contained in a certain countable set, and there are many countable subsets of the reals). I do not think that this is very astonishing.