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Why study set theory?      We like to think that mathematics developed from the need of our ancients to count things. I have four sheep, you have sixteen camels, my tribe has ten dozens of men, you have six hundred wives... etc. etc. But if you look closely, counting how many things you have of a certain type, first required you have be ...


11

Naive set theory: Set theory is the common language to speak about mathematics, so learning set theory means learning the common language. Another aspect is that of counting. Cardinality of sets is a very fundamental notion which be treated naively quite efficiently. Cardinality means counting, so learning set theory means learning to count (beyond the ...


9

Suppose that $\varphi:A\to\wp'(S)$ is a bijection. Let $A_0=\{a\in A:|\varphi(a)|\text{ is even}\}$, and let $A_1=A\setminus A_0$. Then $\{A_0,A_1\}$ is a partition of $A$ into infinite sets, and $A$ is therefore not amorphous. However, there are models of $\mathsf{ZF+\neg AC}$ in which there are amorphous sets.


6

Well, you will always have a proper class of families with a choice function, so in what sense do we mean almost all? Here are a few interpretations. You could say, every family can be extended to one that doesn't have a choice function. But that is really just the failure of $\sf AC$. Fix one counterexample, and use that. Other variants like "...by ...


4

As Andres points out in the comments, $\sf CH$ is not the actual player here. Consider the case that there is an inaccessible cardinal $\kappa$ which is itself a limit of inaccessible cardinals. By forcing we can arrange $\kappa$ to be $\omega_1$. This will certainly say that $\omega_1$ is large in the sense that it is the limit of inaccessible cardinals in ...


3

To expand on Ittay Weiss's post, and maybe this is not exactly what you are looking for, but set theory deals with the mathematical universe. More concrete, set theory has the ability to describe independence results. This, in my opinion, is the real importance in set theory. As Ittay alluded to above, consider $\mathbb{R}$. Does there exist a set $A$ such ...


3

The space $\{0,1\}^\kappa$ has weight $\kappa$ and the required cardinality. For a connected example, that contains all Tychonoff spaces of weight $\kappa$ as a subspace, consider the Tychonoff cubes $[0,1]^\kappa$ which have cardinality $(2^{\aleph_0})^\kappa = 2^\kappa$ as well.


3

Remember the definition of the Cartesian product: $$A\times B=\{\{\{a\},\{a,b\}\}:a\in A,\>b\in B\}.$$ From this, we get $$ \begin{align} \operatorname{trcl}(A\times B)=&(A\times B)\cup\bigcup(A\times B)\cup\bigcup{}^2(A\times B) \cup \operatorname{trcl} \bigcup{}^2(A\times B)\\ =&(A\times B)\cup (\{\{a\}:a\in A\}\cup \{\{a,b\}:a\in A,b\in ...


3

The standard model of Peano Arithmetic -- that is, the usual natural numbers whose Platonic existence most of us believe in instinctively -- is well-founded. However, thanks to Gödel (or just compactness) we know that PA has non-standard models, and a non-standard model of PA is not well-founded. In particular, since arithmetic on elements that can be ...


1

Two comments: There is no such thing as "the" Vitali set. A Vitali set is a set of reals that contains one member from each equivalence class under the equivalence relation $x \sim y \iff x-y \in \mathbb{Q}$. In ZF + DC, every Vitali set has zero inner measure and positive outer measure. If this does not answer your (non existent) question try formulating it ...



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