Firstly, I should say that I came up with this paradox after reading of the Grimm Reapers paradox, but I’m not quite sure how this should be resolved. Nevertheless here is the problem:

Suppose a lady has a countably infinite number of male admirers who all intend to propose to her between $$12:00$$am and $$12:01$$am.

Let each man be named after a positive integer such that each positive integer is the name of corresponding man. So we have man no $$1$$, man no $$2$$, man no $$3$$,...... and so on.

Each man meets and proposes to the lady at exactly $$\left(\frac12\right)^n$$ minutes after $$12:00$$am. Here “n” is the number or name of the man.

The lady will accept the proposal of whichever man proposes to her. When a man proposes - and she accepts the proposal - he places a ring on her finger. A peculiarity of the ring is that inscribed on it is the number of the man who placed it on her finger.

Now whenever a man wants to propose to the lady he checks to see if she is already engaged. That is, he checks for a numbered ring on her finger. If he sees that there is already a ring with a number on her finger then he does not propose anymore. Thus, only a maximum of one man can propose to the lady (since any other man would see a ring on her finger).

Here are my questions:

1. Is there a ring on the lady’s finger by 12:00:30am?
2. If there is then what number is inscribed on the ring?
3. If not then how could no man have proposed to her?

Edit: Someone said I should formalize the statement. I'm not good at using the syntax here so bear me out

Firstly, say Let $$T(n)=12+1/2^n$$ be time of proposal of man number n (note: $$1/2^n$$ is in minutes so this all occurs in time interval (12:00:00, 12:00:30] ). Then let there be a function $$f$$ such that:

1. $$f(n)=1$$ ...............means that the man numbered n proposed (remember that the man number 1 is the last man to approach the lady)
2. $$f(n)=0$$ ...............means that the man numbered n does not propose
3. $$f(n)=1$$ iff $$f(m)=0$$ for all m>n..........means that a man proposes iff no man before him has proposed
• A bit like the one where you put a ball into an urn at time 1/2, then take it out at time 3/4, then put it in at time 7/8, etc - is it in the urn at time 1 or not? I forget the name. Commented Feb 21, 2022 at 11:11
• Yes that sounds like Thomson’s lamp paradox. The difference is: Thomson is asking whether the lamp will be on or off at then end (will the ball be in the urn at the end?) while I am really asking who the lady will be engaged to at the end Commented Feb 21, 2022 at 11:27
• So you're asking for the minimum value of $1 / 2^n$, where $n$ is an integer $> 0$. Well, there is no such minimum, hence your paradox. Commented Feb 21, 2022 at 14:03
• This is one of those times where a prosaic description is not enough. Model the whole process by using mathematical concepts only, i.e define "propose", "assign", the ring's state and the mens' decisions rigorously. Then formulate your question in the same manner. Commented Feb 21, 2022 at 18:37
• @DavidOkogbenin Another perspective is that your process is in my eyes equivalent to the $n$'th man arriving at time $-n \in \{-1, -2, -3, \ldots\}$. So, which man gets to give her the ring? Well, no man arrives first, so the whole process doesn't work. It's like trying to run an algorithm that has no beginning or something like that. Commented Feb 22, 2022 at 15:07

Let's examine the formalization in your edit. I think it's (very close to) an acceptable formalization. Note that the function $$f$$ actually doesn't care about $$T(n)$$, so you don't need $$T$$ in this formalization. Your $$f$$ only remembers that the men come in reverse order in time, which is akin to my comment.

So now, the question "Is there a ring on the lady’s finger by 12:00:30am?", would correspond to "Is $$f(n)=1$$ for any $$n$$?". But the mathematical question that should come before this is "Does such a function $$f$$ exist?". The answer to that is a definite no, so any further questions about $$f$$ are meaningless. In fact, the "paradox" itself can be turned into a proof by contradiction for the non-existence of $$f$$.

As for the informal version, I see essentially two things making the setup ill-defined.

1. You ask about the value of something after an infinite number of steps, or operations. This cannot be supposed to have a well-defined ansewr, unless the sequence is mathematically convergent. This is like Thomson's lamp.

2. This is the heart of the problem. As it is written in words, you basically give a verbal proposed definition of variables corresponding to whether each man proposes or not. As becomes apparent with your formalization using $$f$$, no outcome follows your rules, so it is senselense to ask about the properties of the outcome. (Well, one may reason mathematically about properties of purported but non-existent things, but that's like asking "how loud is a purple dog?").

I think the moral of the story should be that a mathematicians first questions should be, is what we're talking about well-defined? And does it exist? This should come before any other questions, and resolves the paradox with a big "nope".

As for all your questions in the comments, basically there's nothing wrong with countably (or even uncountably) many assignments, or having their actions depend on previous men, or anything else really in and of itself. But let's say we have freedom over the arrival times. What we can say is that the process can occur (meaning that the function $$f$$ exists) if and only if there is a man arriving before all others (meaning that $$\exists n\forall m\ne n: T(n) < T(m)$$), at least assuming distinct arrival times. So to answer your question of wherein the mathematical contradiction lies, it truly is nothing more and nothing less than the set of arrival times having no minimum.

• oh interesting. As I understand it, from the 3 properties of $f$ that i mentioned earlier let us consider the following statements: (1) $f(n)=1$ if $f(m)=0$ for all $m>n$; (2) $f(n)=1$ only if $f(m)=0$ for all $m>n$; (3) $∃n∀m≠n:T(n)<T(m)$. Note: that i simply split the third property of $f$ in my formalization to get (1) and (2). Now it seems to me that any two of the three statements i just mentioned can be true. However, i do not believe all three can be true. Am i understanding things correctly? Commented Feb 22, 2022 at 22:53
• Oh sorry, my (3) statement was a copy and paste of your statements. I don’t know how to use the math formula syntax on this site. My (3) statement was meant to be “There is no n, such that $T(n)<T(m), for all m Commented Feb 23, 2022 at 6:05 • @DavidOkogbenin Almost; the$m>n$should be replaced by$T(m)<T(n)$. But then you're right, and we can even characterize the possible functions in each case: 1+2:$f$has a single$1$at the first arrival. 2+3:$f$has only$0$'s, or a single$1$that can be placed anywhere. 1+3:$f$has infinitely many$1$'s, in such a way that for all$n$there is an$m$with$T(m)<T(n)$and$f(m)=1$. Note also that we made no assumptions on the number of men, they can be finite, countable or uncountable with no changes to these statements. (Except that the finite case of course makes (3) impossible). Commented Feb 23, 2022 at 9:00 Here's a simple proof of the comments based on contradiction. Assume that man $$n$$ gave her the ring. That would mean she didn't have ring on from time midnight to midnight +$$\frac 1 n$$ minutes on. But then man $$n+1$$ should have been able to put the ring on at time midnight + $$\frac 1 {n+1}$$. Thus for no natural number $$n$$ can the $$n$$th man be the right one. • So does that mean that the set up is impossible? All the men cannot have such a proposal strategy? Commented Feb 22, 2022 at 15:44 • The setup is impossible because you can't have a countably infinite number of men :). Putting it in abstract math terms is asking what is the minimum value of$\frac 1 n$, which does not exist – Alan Commented Feb 22, 2022 at 15:45 • Sorry, why do you say you cannot have a countably infinite number of men? Why can one not have a set with a countably infinite number of items? I agree, there is no minimum value of$1/n\$ Commented Feb 22, 2022 at 15:54
• @DavidOkogbenin I was referring to the real world situation. You can have an infinite set, but I'm pretty sure the universe is not big enough to hold an infinite number of men.
– Alan
Commented Feb 22, 2022 at 16:15
• Mathematically you cannot do it in a finitary way since at each finite number you need to check the infinite amount of people who go before the person. So there is no way to approach this with finite intuition. Thus we are left with the purely infinite approach, which is saying the engagement happens at a minimum which doesn't exist.
– Alan
Commented Feb 22, 2022 at 16:23