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29

Well. That depends on whom you might ask this. Set theory might be inconsistent. In particular $\sf ZFC$ and its extension by large cardinal axioms. It's a nontrivial thing, to feel safe with these theories, and it takes a lot of practice and time until you understand that $\sf ZFC$ is self-evident to some extent, and [some] large cardinal axioms are ...


7

If you're looking for motivation to pursue other foundations, I recommend the article "Rethinking Set Theory" by Tom Leinster: http://arxiv.org/pdf/1212.6543v1.pdf. In particular, he's providing a gentle, well-motivated introduction to Lawvere's Elementary Theory of the Category of Sets (ETCS). Leinster mentions at least two complaints. First, in ...


6

Recall that a tree is a partial order that for each point $x$, the cone below $x$ is well-ordered. An infinite tree is a tree with infinitely many nodes. Sometimes, however it is easier to work with "reversed" trees. With reversed ordering, being well-founded is equivalent to not having an infinite branch. And that is something which comes up often in ...


5

As was brought up in the comments, there is actually a counterexample. I'll give a simpler one than I gave in the comments, with the same idea. Take $B = P(\{1,2,3\})$, and $$A = \{\emptyset,\{1\},\{2,3\},\{1,2,3\}\}.$$ Then $A$ has the automorphism which switches $\{1\}$ and $\{2,3\}$. But this can't extend to an automorphism of $B$, since every ...


3

Typically, in set theory language, a (possibly infinite) "tree" means a nonempty set of sequences of symbols (for example integers), which is closed under taking prefixes. What does this have to do with the graph-theoretic concept of a tree? Assume that you're labeling all edges in the tree with a symbol such that no node has two outgoing edges with the ...


2

As asked, you actually need to specify a little bit more to be precise. I'll first assume you're asking about infinite, binary branching trees. A binary branching tree $T$ is a collection of finite strings of zeros and ones, called nodes, such that, whenever a string $\sigma$ is in $T$, all of its prefixes are in $T$. An infinite binary branching tree is a ...


2

This is a nice question and the answer is yes. The key is the following result of Solovay: if $G,H$ are mutually generic for $\mathbb{P}$ (meaning that $G\times H$ is generic for $\mathbb{P}\times\mathbb{P}$) then $V[G]\cap V[H]=V$. You can find a proof in this MO answer (your own argument was basically going in this direction). So let $\tau$ be a name ...


1

In order to have a Suslin scheme, you must have a set $A_\sigma$ for each $\sigma\in{^{<\omega}X}$; since you have only $A_\sigma$ for $|\sigma|\le 2$, you don’t have a Suslin scheme and cannot define the result of the $\mathscr{A}$-operation. If you set $A_\sigma=\varnothing$ for all $\sigma\in{^{<\omega}X}$ of length greater than $1$, so as to have a ...



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