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Aaron Meyerowitz
Aaron Meyerowitz
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The counter-intuitiveness of the axiom of choice is clouding the real issue here ( I think).

Suppose there If I have a secret number and you try to guess a larger one, what are your odds of success? is an fixed but unknownthere a way to make them better (if you sequence of reals $x_0,x_1,\cdots$have $99$ friends). Furthermore, the referee assures you that all butYou have only finitely many are $0$ (no assumption on howways to fail and infinitely many are non-zero nor which)to succeed, but we can't really assign a probability. You may chooseThat is also behind this puzzle

Going back to the given problem, suppose that each box has a guard. At a certain time each guard sees the contents of every other box then guesses the contents of their own. Under AoC with an agreed set of terms to eliminate (evenequivalence class representatives and the familiar protocol we know that all but one) and examine their valuesfinitely many will be correct. Then you choose some other termThat is strange but we are used to this consequence of the axiom of choice. If it turns outyou have full knowledge of all this and are required to bebecome the guard for a box of your choice $0$(after looking into as many other boxes as you win. It seemswish) can we say that your chancesyou are good (even infinitely good in some sensecertain to make a correct guess?) but you gain nothing by what you see Not really.

Now suppose that there are $100$ people who play this game onThe situation in the same sequence of reals. They can make a strategyquestion above seems clearer to me if we stipulate ahead of time that all but then not communicatefinitely many boxes contain $0$ (no assumption , other than finiteness on which the exceptions are). Can we say that If there is a strategy which makes themare totally$100$ potential guards who are friends, can they be sure that each can pick a box and guess $0$ (what else can they do?) with at least $99$ will get a term equal to $0$correct? Perhaps. As aboveyes, putas before split the terms inboxes into $100$ infinite subsequences. Person, person $k$ will looklooks at all the valuesboxes in every subsequence except subsequence $k$,other than his own then choose from subsequence $k$chooses a termbox further along than anythe last non-zero term from anybox in every other subsequence.

But ifIf you are personare one member of this set of $17$ can we say that$100$ friends are your chanceodds of getting a non-zero value is lower than ifbeing correct (when you were the only person playing? Maybeguess $\frac{0}{100}=0$. Otherwise it doesn't$0$) improved $100$ fold, or at all? Not really matter.

To connect this with the given problem, assume again an arbitrary real sequenceEssentially $100$ people are assigned hidden integers and youmust name an integer greater than their assigned integer (perhaps as one member of a group) needso only finitely ways to deduce one value based on some or all of the rest. There is no harm in revealingfail and infinitely many to the referee (after the sequence has been setsucceed!) the family of coset representatives and the entire strategy. And then the refereeIf each person can go back to the sequence (or $100$ subsequences) and replaces the values bysee all the difference between their actual and "predicted" values. It shouldn't matterother integers then only one will fail.

In other words, given AoC, we may assume a known set of coset representatives such that every real sequence is is (uniquely) the termwise sum of a coset representative and a sequence with only finitely many non-zero terms. Once we accept that (which is weird but familiar) we may assume wlogthere is no loss in saying that the all zerochosen sequence is in a certain coset representative and that(so why not the chosencoset of the zero sequence is from that equivalence class.)?

The counter-intuitiveness of the axiom of choice is clouding the real issue here ( I think).

Suppose there are is an fixed but unknown to you sequence of reals $x_0,x_1,\cdots$ . Furthermore, the referee assures you that all but finitely many are $0$ (no assumption on how many are non-zero nor which). You may choose a set of terms to eliminate (even all but one) and examine their values. Then you choose some other term. If it turns out to be $0$ you win. It seems that your chances are good (even infinitely good in some sense?) but you gain nothing by what you see.

Now suppose that there are $100$ people who play this game on the same sequence of reals. They can make a strategy ahead of time but then not communicate. Can we say that there is a strategy which makes them totally sure that at least $99$ will get a term equal to $0$? Perhaps. As above, put the terms in $100$ infinite subsequences. Person $k$ will look at all the values in every subsequence except subsequence $k$, then choose from subsequence $k$ a term further along than any non-zero term from any other subsequence.

But if you are person $17$ can we say that your chance of getting a non-zero value is lower than if you were the only person playing? Maybe $\frac{0}{100}=0$. Otherwise it doesn't really matter.

To connect this with the given problem, assume again an arbitrary real sequence and you (perhaps as one member of a group) need to deduce one value based on some or all of the rest. There is no harm in revealing to the referee (after the sequence has been set) the family of coset representatives and the entire strategy. And then the referee can go back to the sequence (or $100$ subsequences) and replaces the values by the difference between their actual and "predicted" values. It shouldn't matter.

In other words, given AoC, we may assume a known set of coset representatives such that every real sequence is is (uniquely) the termwise sum of a coset representative and a sequence with only finitely many non-zero terms. Once we accept that (which is weird but familiar) we may assume wlog that the all zero sequence is a coset representative and that the chosen sequence is from that equivalence class.

The counter-intuitiveness of the axiom of choice is clouding the real issue here ( I think). If I have a secret number and you try to guess a larger one, what are your odds of success? is there a way to make them better (if you have $99$ friends). You have only finitely many ways to fail and infinitely many to succeed, but we can't really assign a probability. That is also behind this puzzle

Going back to the given problem, suppose that each box has a guard. At a certain time each guard sees the contents of every other box then guesses the contents of their own. Under AoC with an agreed set of equivalence class representatives and the familiar protocol we know that all but finitely many will be correct. That is strange but we are used to this consequence of the axiom of choice. If you have full knowledge of all this and are required to become the guard for a box of your choice (after looking into as many other boxes as you wish) can we say that you are certain to make a correct guess? Not really.

The situation in the question above seems clearer to me if we stipulate ahead of time that all but finitely many boxes contain $0$ (no assumption , other than finiteness on which the exceptions are). If there are $100$ potential guards who are friends, can they be sure that each can pick a box and guess $0$ (what else can they do?) with at least $99$ correct? yes, as before split the boxes into $100$ subsequences, person $k$ looks at all the boxes in every subsequence other than his own then chooses a box further than the last non-zero box in every other subsequence.

If you are one member of this set of $100$ friends are your odds of being correct (when you guess $0$) improved $100$ fold, or at all? Not really.

Essentially $100$ people are assigned hidden integers and must name an integer greater than their assigned integer (so only finitely ways to fail and infinitely many to succeed!). If each person can see all the other integers then only one will fail.

In other words, given AoC, we may assume a known set of coset representatives such that every real sequence is is (uniquely) the termwise sum of a coset representative and a sequence with only finitely many non-zero terms. Once we accept that (which is weird but familiar) there is no loss in saying that the chosen sequence is in a certain coset (so why not the coset of the zero sequence)?

Source Link
Aaron Meyerowitz
Aaron Meyerowitz
  • 30.1k
  • 1
  • 48
  • 104

The counter-intuitiveness of the axiom of choice is clouding the real issue here ( I think).

Suppose there are is an fixed but unknown to you sequence of reals $x_0,x_1,\cdots$ . Furthermore, the referee assures you that all but finitely many are $0$ (no assumption on how many are non-zero nor which). You may choose a set of terms to eliminate (even all but one) and examine their values. Then you choose some other term. If it turns out to be $0$ you win. It seems that your chances are good (even infinitely good in some sense?) but you gain nothing by what you see.

Now suppose that there are $100$ people who play this game on the same sequence of reals. They can make a strategy ahead of time but then not communicate. Can we say that there is a strategy which makes them totally sure that at least $99$ will get a term equal to $0$? Perhaps. As above, put the terms in $100$ infinite subsequences. Person $k$ will look at all the values in every subsequence except subsequence $k$, then choose from subsequence $k$ a term further along than any non-zero term from any other subsequence.

But if you are person $17$ can we say that your chance of getting a non-zero value is lower than if you were the only person playing? Maybe $\frac{0}{100}=0$. Otherwise it doesn't really matter.

To connect this with the given problem, assume again an arbitrary real sequence and you (perhaps as one member of a group) need to deduce one value based on some or all of the rest. There is no harm in revealing to the referee (after the sequence has been set) the family of coset representatives and the entire strategy. And then the referee can go back to the sequence (or $100$ subsequences) and replaces the values by the difference between their actual and "predicted" values. It shouldn't matter.

In other words, given AoC, we may assume a known set of coset representatives such that every real sequence is is (uniquely) the termwise sum of a coset representative and a sequence with only finitely many non-zero terms. Once we accept that (which is weird but familiar) we may assume wlog that the all zero sequence is a coset representative and that the chosen sequence is from that equivalence class.