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April 27th, 2014, 11:02 AM   #1
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Set Theory

I'm reading Jech's book. Can someone solve exercises
1.6.
1.7.
2.12
help me to understand lemmae 3.6.-3.10. and 5.2.
theorem
3.11.
4.5.
4.8.-Baire category theorem
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April 28th, 2014, 11:00 AM   #2
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({} є S and (for each x є S) x U {x} є S)

We call a set S with the above property inductive.

A set T is transitive if x є T implies x subset of T.

Exercise 1.6. If X is inductive, then {x є X: x is transitive and every nonempty z subset of x has an є-minimal element} is inductive (t is є-minimal in z if there is no s є z such that s є t)

Last edited by raul14; April 28th, 2014 at 11:04 AM.
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May 22nd, 2014, 12:27 PM   #3
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Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology
Theorem 4.5.
cardin
Every perfect set has cardinality of $\displaystyle R$.

Proof. Given a perfect set $\displaystyle P$, we want to find a one-to-one function $\displaystyle F$ from $\displaystyle {0, 1}^\omega$ into $\displaystyle P$.

$\displaystyle {0, 1}^\omega$ is equipotent to $\displaystyle {0, 2}^\omega$, given a sequence of 0s and 1s, you just map it to the sequence with 2s instead of 1s.

C (the Cantor set) is equipotent to $\displaystyle {0, 2}^\omega$, given a real of the form $\displaystyle \sum_{n=1}^{\infty}\frac{a_n}{3^n}$, where each $\displaystyle a_n$= 0 or 2, map it to the sequence with coresponding 0s and 2s, and so the cardinality of C = $\displaystyle 2^\aleph_0$.
Therefore cardinality of $\displaystyle R$>=$\displaystyle 2^\aleph_0$, because C is a subset of reals.

As the set $\displaystyle Q$ is dense in reals, every real number r is equal to sup{q element of Q: q<r} and because Q is countable, it follows cardinality of $\displaystyle R$<=cardinality of power set of Q = $\displaystyle 2^\aleph_0$

I don't understand the last paragraph. Help?

By Cantor Bernstein cardinality of $\displaystyle R$=$\displaystyle 2^\aleph_0$
So, R is equipotent to C, and to $\displaystyle {0, 1}^\omega$.
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