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This is a past exam exercise I'm struggling to solve: I was able to proove $Q\cup f[\mathcal{T}]\subseteq \mathcal{T}$ and had no luck afterwards.

We have:

  1. $A'\subseteq B \subseteq A$ such that $A=_c A'$
  2. $f:A \rightarrow A'$ an isomorphism.
  3. $Q=B / f[A]$
  4. $\mathcal{T}=\left\{X\subseteq A| Q\cup f[X]\subseteq X\right\}$ and $T=\bigcap_{X\in \mathcal{T}}X$.

I would like to prove the following:

  1. $Q\cup f[T]=T$

Any hints on the $T\subseteq Q \cup f[T]$?

Thank you in advance for your time and effort. This is part of a guided proof of Schröder-Bernstein Theorem.

dfeuer
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epsilon
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  • What $B/f[A]$ means? – Emanuele Paolini Oct 21 '13 at 06:26
  • $B-f[A]$ which since $f$ is an isomorphism equals to $B-A'$. – epsilon Oct 21 '13 at 06:32
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    Brian M. Scott once gave a complete proof of the Cantor-Bernstein theorem on this site. I find that giving elaborated proofs for a theorem (whose proof is not very short) that you can find in many books, websites and notes is by definition very broad. Please restrict to a single point which you don't understand in the process of the proof, and I will retract my vote to close/vote to reopen. As it stands, the question is too broad. – Asaf Karagila Oct 21 '13 at 07:02
  • By the way... nice title on your webpage. Although I thought you set theorists always have a choice! – epsilon Oct 21 '13 at 07:13
  • Any hints on the first 3 questions is more than enough. If someone knows a similar proof of the theorem that would also be great. – epsilon Oct 21 '13 at 07:33
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    @Asaf: Complete proof here, and here with diagrams. – Brian M. Scott Oct 22 '13 at 02:27
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    Formal proof (in excruciating detail) at http://dcproof.com/CBS.htm and http://dcproof.com/KnasterFPL.htm – Dan Christensen Oct 22 '13 at 04:47
  • Thank you all very much for helping out. – epsilon Oct 22 '13 at 06:22

1 Answers1

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We have $f[T]\subseteq T$, so $f[f[T]]\subseteq f[T]$. We also have $Q\subseteq T$ and therefore $f[Q]\subseteq f[T]$, so $f[Q\cup f[T]]\subseteq Q\cup f[T]$, and hence $Q\cup f[T]\in\mathscr{T}$. The desired inclusion now follows immediately.

Brian M. Scott
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