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April 27th, 2014, 11:02 AM  #1 
Newbie Joined: Apr 2014 From: zagreb Posts: 5 Thanks: 0  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 
April 28th, 2014, 11:00 AM  #2 
Newbie Joined: Apr 2014 From: zagreb Posts: 5 Thanks: 0 
({} 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 єminimal element} is inductive (t is єminimal 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. 
May 22nd, 2014, 12:27 PM  #3 
Senior Member Joined: Apr 2014 From: zagreb, croatia Posts: 234 Thanks: 33 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 onetoone 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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