I Had To Reverse The Order Of "23 Yo" And "libertarian Socialist" In My Bio Because On My Phone It Would

And they all have an entire wall dedicated to funko pops, cruelty squad style.

It's so funny how the game store used to sell games the music store used to sell music and hot topic used to sell goffic accessories but now they all stock the same anime backpacks and novelty fnaf soap of whatever

My magnum opus!

j aliens

Hi everypony. This is my ponysona, Limestone River.

Here's another weird model of ZFC relating to the axiom of regularity that fucked me up when I learned about it. The axiom of regularity implies that there can be no set {x_n | n ∈ ω} such that x_{n+1} ∈ x_n for each n ∈ ω.

Let's construct our model. Add to the language of ZFC countably infinitely many constants c_0, c_1, c_2, ... and let Γ be the set of sentences Γ = {c_{n+1} ∈ c_n | n ∈ ω}. We will use the compactness theorem to show that there is a model of ZFC ∪ Γ.

Let Δ be a finite subset of Γ and Let J be any model of ZFC. Since Δ is finite, there is a maximum k such that the sentence c_k ∈ c_{k-1} is in Γ. Add to J the definitions, for each n ≤ k, c_nJ = k - n, and for each n > k, set c_nJ = 0. Then for all 1 < n ≤ k, c_n = k-n ∈ k-n+1 = c_{n-1}, and so J is a model of ZFC ∪ Δ.

Thus, by the compactness theorem, there exists a model of ZFC ∪ Γ.

This is very surprising, and at first glance seems to contradict the axiom of regularity! But what it really means is that the sets x_n from the first paragraph can exist, but they cannot be gathered together in a set.

downward lowenheim-skolem is so fucked up to me. what do you *mean* there's a countable model of first-order set theory

Jammin out with twilight sparklt

:-)

Some pony doodles