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karthickgururaj

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karthickgururaj last won the day on November 3 2014

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About karthickgururaj

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  1. Question: what happens when there are only 2 bankers left?
  2. For a finite set, like A = {1, 2, 3}, clearly the super set has higher number of elements (in this case 8), so we can't define an injection from the superset to the original set. For an infinite set, like N = {1, 2, 3, .. }, it is not immediately clear whether we can (or not) define the injection from the superset of N (lets call it S(N)) to N. (So, I have just restated part of the OP!)
  3. Interesting! Thanks for the link. Reading the wiki article exposes one assumption in my solution. I assumed that, while the super set has infinite number of elements, each individual element of the super set has a finite number of elements. In other words, I assumed that (for example), "the set of all even numbers" is NOT a member of super set. If that assumption is removed, my solution won't work.
  4. For finite sets it is easy to see that a bijection exists, if there are two injections, one from A to B, and other from B to A. If f: A -> B is an injection from A to B and g: B -> A is an other injection from B to A, then the number of elements in A and B are both same. So, a bijection can be defined from A to B. I googled a bit on this and read about Schröder–Bernstein theorem, which essentially asserts that if there are two injections in reverse directions, then a bijection exists.
  5. karthikgururaj proposes two injections (both of which are far from being surjections also) rather than a single bijection. So the question is, are two (one each direction) injections equivalent to a bijection? I thought about this when posting the solution.. In truth, it doesn't appear to be an equivalent to bijection
  6. I think I got this.. pleased as punch
  7. I agree - I was wrong in my earlier post. The diagram in particular has a big mistake.
  8. I think with that hint and plasmid's earlier work, the solution seems easy.. I take back the second statement
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