The Engagement paradox

Question

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

Answer 1

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 $f$.

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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?").

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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.

Answer 2

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.