This article is what happens when you teach teenagers more complicated mathematics without any clear everyday real-world applications. The mathematics of this "proof" only make sense in an infinite amount of worlds different from ours. If you are an aspiring False Shepard, or the Lamb, it may be applicable to you! Also RUN! Comstock maybe after you!
Features HEAVY SPOILERS for Bioshock Infinite.
Bioshock Infinite flirts with the theme of … you guessed it, infinity. The main assertion of the game being, that the multiverse is real (in the context of the game)
At the end of the plot, a group of Elizabeth drown Booker DeWitt before he could accept baptism and be reborn as Zachary Hale Comstock.
This is to say that Booker’s baptism represents a node on a tree. In this case, a binary choice. Either accept the baptism, and found Colombia, or deny, remain Booker DeWitt, and gamble away his child Anna.
It is an interesting concept however, what actually states that this is the only node which might lead to Colombia and Comstock? Is Booker Dewiit really that special that his choices split the universes? Because if so, great. Otherwise the plot falls apart. (Though Elizabeth still causes a paradox)
Then Burial at Sea happens. There’s still at least one Booker DeWitt left, in the city of Rapture, where else of course. So that was not the only node which mattered to this. Which means that the amount of universes is … might as well call it infinite. The game’s title already gave us that.
We can make assertions regarding infinite sets using Mathematical Induction.
$${\displaystyle \forall P\,{\Bigl (}P(0)\land \forall k{\bigl (}P(k)\to P(k+1){\bigr )}\to \forall n\,{\bigl (}P(n){\bigr )}{\Bigr )},} $$
The gist is that if we can prove the assertion for n = 1, and assume that it is true for n, then we can algebraically prove that it is true for all n + 1. This normally applies to non-negative integers. We’re throwing that right out of the window and applying it to something as undefined as a world.
To truly eliminate Booker DeWitt, the game has to prove that for each world, it can remove its Comstock / Booker from its world.
The assertion, for the base game, works. At a double rate. A booker and a comstock are present, and both are deceased by the end of the game. The assertion is true for n = 1. In the Burial at Sea DLC however, Elizabeth searches up a Booker who is alive in the 50s! If the baptism scene would have been enough to eliminate a line of Bookers and Comstocks, then this is a clear demonstration that it failed. n = 2 is false, because an out-of-world Elizabeth has to deal with him, as this world's Booker's child has unfortunately passed away.
$$ \begin{align*} \sum_{i=1}^{\infty} |Booker_i|=0 \end{align*} $$
Not every world deals with its own Comstock / Booker, therefore we cannot assert the totality of Comstocks and Bookers can be erased. In fact maybe even you are one!
I’m sorry if you expected clean, formal mathematics, maybe another day.