Thanks for clearing up the countability. It’s clear that there are some cases where taking limits will fail (like when the utility is discontinuous at infinity), but I don’t have an intuition about how that issue is related to countability.
No, I mean a function whose limit doesn’t equal its defined value at infinity. As a trivial example, I could define a utility function to be 1 for all real numbers in [-inf,+inf) and 0 for +inf. The function could never actually be evaluated at infinity, so I’m not sure what it would mean, but I couldn’t claim that the limit was giving me the “correct” answer.
The function could never actually be evaluated at infinity, so I’m not sure what it would mean, but I couldn’t claim that the limit was giving me the “correct” answer.
If you accept the Axiom of Infinity, there’s no problem at evaluating a function at infinity. The problem is rather that omega is a regular limit cardinal, so there’s no way to define the value at infinity from the value at the successor, unless you include in the definition an explicit step for limit cardinals. You can very well define a function that has 1 as value on 0 and on every successor cardinal, but 0 on every limit cardinal. The function will indeed be discontinuous, but its value at omega will be perfectly defined (I just did).
The problem with saying a function is not continuous at infinity is that the definitions of ‘continuous’ requires the standard definition of ‘limit’ (sigma-epsilon), while the definition of limits at infinity uses the same nomenclature and similar notation, but expresses something different.
Consider the case where F(X-epsilon) is 1, F(X) is 0, and F(X+epsilon) is either 0 or undefined. The common thought there is that the limit at X does not exist; why is that any different just because X is infinite, without sacrificing the concept which allows us to talk about continuity in terms of limits?
The problem with saying a function is not continuous at infinity is that the definitions of ‘continuous’ requires the standard definition of ‘limit’ (sigma-epsilon), while the definition of limits at infinity uses the same nomenclature and similar notation, but expresses something different.
Well… I guess you can see it that way if you want, but in set theory (again, everything I say is under the axiom of infinity) both notions are unified under the notion of limit in the order topology. In this way, you can define a continuous function for every transfinite ordinal.
Consider the case where F(X-epsilon) is 1, F(X) is 0, and F(X+epsilon) is either 0 or undefined. The common thought there is that the limit at X does not exist; why is that any different just because X is infinite, without sacrificing the concept which allows us to talk about continuity in terms of limits?
Yes, I understand that the concept of limit in calculus and set theory means something a little different. Possibly this is just arguing over definitions: in calculus, it is said that a limit doesn’t exist when the function has a different value w.r.t. the value calculated using the topology of its domain, but in set theory a limit is defined in a different way, using only the order topology. In this sense, a function can be defined at omega, at omega+1, omega+2, etc. After all, that’s the raison d’etre of the entire concept. Under this assumption, you just say that if the function is defined at omega, and if it has a different value at omega than the one defined from its order topology, you just say that it’s discontinuous. Let me clarify with an example: the function y = 2x, defined in set theory, would have at omega the limit omega (demonstration below). If you just define a similar function but that has at omega the value 0, then you have a discontinuous function, because the (topological) limit is different from the defined value. Of course, if you want, you can just say that the (simple) limit doesn’t exists when this situation arise, I was on the other side pointing to the fact that a function can be perfectly defined and continuous or discontinuous at infinity.
Thesis Omega is the limit of y=2x, defined on the natural number. Demonstration The intersection of the range of the function with omega is still the range of the function. The range is unbounded but every one of its member is finite: since omega is regular limit, it is never reached by the sequence. So, any ordinal greater then or equal to omega is an upper bound. But omega is also an initial ordinal, so is the least upper bound. That is, the least upper bound of the intersection of the range of the function with omega is still omega, so by definition omega is the limit of the range of the function. QED.
TIL that mathematicians commonly strictly define terms in one context, then extend them into other contexts in ways that are not strictly compatible with the original context.
Now I understand that two people with significantly different levels of math education and understanding lack the common vocabulary required to trivially communicate basic math-related concepts.
It’s still going to be hard to convince me that the sum of an infinite number of days, each of which has infinite positive utility and finite negative utility, will ever be lower than zero.
It’s still going to be hard to convince me that the sum of an infinite number of days, each of which has infinite positive utility and finite negative utility, will ever be lower than zero.
Remember you’re not allowed to talk about infinity except as a limit.
Clearly the limit of this sequence is the sum you’re talking about (that is, the utility of an infinite number of immortal people who start in sphere A, where we move one each day to sphere B). At the same time, clearly the limit of this sequence is negative.
(Of course the “real” answer is that it’s not well defined; there are many sequences we can construct that come out as “infinite immortal people” in the limit, and the utility is different depending which we pick. But this is an example of why “lower than zero” is as legitimate an answer as “higher than zero”).
Except that in the original problem, there cannot be more days than people;
...each day, one more person gets permanently transferred across to... (Emphasis added.)
Then again, rephrasing the problem in equivalent ways has interesting effects:
Assume that the number of people is countable; assign each of them a natural number, but don’t tell them which one it is.
Suppose that on the Nth day you move the person with the Nth prime number to the opposite sphere; every individual prefers case 1, where they have 100% chance of infinite happiness.
Suppose that you tell everyone their number and on the Nth day you move the person with the Nth number to the opposite sphere; every individual prefers case 2, where they have a finite period of agony followed by infinite happiness.
What’s the Erdos number of an infinite number of monkeys juggling an infinite number of bananas?
Um… -inf and +inf are not real numbers. (Noting that your function as described is undefined at -inf.)
In addition, the definition of continuous restricts it to points which exist on an open interval; if the limit from below and limit from above are equal to the value at X, then the function is continuous on an open interval containing X. How do you determine the limit as X approaches +inf from above?
Thanks for clearing up the countability. It’s clear that there are some cases where taking limits will fail (like when the utility is discontinuous at infinity), but I don’t have an intuition about how that issue is related to countability.
You said ‘discontinuous at infinity’. Did you mean ‘the infinite limit diverges or otherwise does not exist’?
No, I mean a function whose limit doesn’t equal its defined value at infinity. As a trivial example, I could define a utility function to be 1 for all real numbers in [-inf,+inf) and 0 for +inf. The function could never actually be evaluated at infinity, so I’m not sure what it would mean, but I couldn’t claim that the limit was giving me the “correct” answer.
If you accept the Axiom of Infinity, there’s no problem at evaluating a function at infinity. The problem is rather that omega is a regular limit cardinal, so there’s no way to define the value at infinity from the value at the successor, unless you include in the definition an explicit step for limit cardinals.
You can very well define a function that has 1 as value on 0 and on every successor cardinal, but 0 on every limit cardinal. The function will indeed be discontinuous, but its value at omega will be perfectly defined (I just did).
The problem with saying a function is not continuous at infinity is that the definitions of ‘continuous’ requires the standard definition of ‘limit’ (sigma-epsilon), while the definition of limits at infinity uses the same nomenclature and similar notation, but expresses something different.
Consider the case where F(X-epsilon) is 1, F(X) is 0, and F(X+epsilon) is either 0 or undefined. The common thought there is that the limit at X does not exist; why is that any different just because X is infinite, without sacrificing the concept which allows us to talk about continuity in terms of limits?
Well… I guess you can see it that way if you want, but in set theory (again, everything I say is under the axiom of infinity) both notions are unified under the notion of limit in the order topology.
In this way, you can define a continuous function for every transfinite ordinal.
Yes, I understand that the concept of limit in calculus and set theory means something a little different. Possibly this is just arguing over definitions: in calculus, it is said that a limit doesn’t exist when the function has a different value w.r.t. the value calculated using the topology of its domain, but in set theory a limit is defined in a different way, using only the order topology. In this sense, a function can be defined at omega, at omega+1, omega+2, etc. After all, that’s the raison d’etre of the entire concept. Under this assumption, you just say that if the function is defined at omega, and if it has a different value at omega than the one defined from its order topology, you just say that it’s discontinuous.
Let me clarify with an example: the function y = 2x, defined in set theory, would have at omega the limit omega (demonstration below). If you just define a similar function but that has at omega the value 0, then you have a discontinuous function, because the (topological) limit is different from the defined value.
Of course, if you want, you can just say that the (simple) limit doesn’t exists when this situation arise, I was on the other side pointing to the fact that a function can be perfectly defined and continuous or discontinuous at infinity.
Thesis
Omega is the limit of y=2x, defined on the natural number.
Demonstration
The intersection of the range of the function with omega is still the range of the function.
The range is unbounded but every one of its member is finite: since omega is regular limit, it is never reached by the sequence. So, any ordinal greater then or equal to omega is an upper bound. But omega is also an initial ordinal, so is the least upper bound.
That is, the least upper bound of the intersection of the range of the function with omega is still omega, so by definition omega is the limit of the range of the function. QED.
TIL that mathematicians commonly strictly define terms in one context, then extend them into other contexts in ways that are not strictly compatible with the original context.
Now I understand that two people with significantly different levels of math education and understanding lack the common vocabulary required to trivially communicate basic math-related concepts.
It’s still going to be hard to convince me that the sum of an infinite number of days, each of which has infinite positive utility and finite negative utility, will ever be lower than zero.
Remember you’re not allowed to talk about infinity except as a limit.
Consider the sequence:
etc.
Clearly the limit of this sequence is the sum you’re talking about (that is, the utility of an infinite number of immortal people who start in sphere A, where we move one each day to sphere B). At the same time, clearly the limit of this sequence is negative.
(Of course the “real” answer is that it’s not well defined; there are many sequences we can construct that come out as “infinite immortal people” in the limit, and the utility is different depending which we pick. But this is an example of why “lower than zero” is as legitimate an answer as “higher than zero”).
Except that in the original problem, there cannot be more days than people;
Then again, rephrasing the problem in equivalent ways has interesting effects:
Assume that the number of people is countable; assign each of them a natural number, but don’t tell them which one it is.
Suppose that on the Nth day you move the person with the Nth prime number to the opposite sphere; every individual prefers case 1, where they have 100% chance of infinite happiness.
Suppose that you tell everyone their number and on the Nth day you move the person with the Nth number to the opposite sphere; every individual prefers case 2, where they have a finite period of agony followed by infinite happiness.
What’s the Erdos number of an infinite number of monkeys juggling an infinite number of bananas?
Um… -inf and +inf are not real numbers. (Noting that your function as described is undefined at -inf.)
In addition, the definition of continuous restricts it to points which exist on an open interval; if the limit from below and limit from above are equal to the value at X, then the function is continuous on an open interval containing X. How do you determine the limit as X approaches +inf from above?
MrMind explains in better language below.