Breaking Bad Redux.

nonexample ,Update from 10 months ago. Deepseek just nailed the same question.

Does there exist an infinite abelian group such that each proper subgroup is finite?

(10 months ago) Hey Deepseek! Does there exist an infinite abelian group such that each proper subgroup is finite?

Deepseek. No...

In reality:

For an example consider a colimit indexed by the natural numbers, of cyclic groups of order 2^n, under monomorphic bonding maps. (Prufer group).

Strong Floor, No Ceiling, Can't Leak!

S: In a Manger (Ippolitov-Ivanov)( Civ V)
A: Fairytale of New York (The Pogues)
B: Run Rudolph Run (Chuck Berry)
C: Please Come Home for Christmas (Don Henley)
D: Santa Claus is Coming to Town (Bruce Springsteen)
F: I saw Mommy Kissing Santa Claus

Watch it cold.

Lots to like, the music alone is enough.

This looks great! Nice collection of topics and you can't beat the price.

Lol. Sorry for not seeing the joke.

Johnny Cash reference. All sorted hopefully.

``...suspended for 72 hours for expressing a desire to shoot the author of an article. The post, made 11/10, stated: "I want to shoot the author of this article just to watch him die." 1/2''

Needed that no retract of X is homotopy equivalent to the infinite earring.

Helpful if small open sets of X have contractible components.

Is pi1(X) guaranteed to be Cauchy Complete?

Dubious example. Free group in the sense of Markov over a T2 space should be Cauchy complete.

What's really going on here?

Here by compactly generated, we mean has a compact set of generators.

Intuition says Player B, since B starts with 8 rounds of guaranteed win rather than 4.

B is player A, if A had the best possible outcome on the first 4 hands.

So kind of a dominance argument.

Nice! And not just any old bonding maps.

If I'm understanding the pictures,

Your construction also arranges that each point preimage of each bonding map is a Peano continuum.

4-4. Bottom 9th.