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Can infinite sets differ finitely?

  1. Oct 3, 2008 #1
    Can two infinite sets differ by a finite number of elements?
     
  2. jcsd
  3. Oct 3, 2008 #2
    Take for example [itex]A = \mathbb{N}[/itex] and [itex]B = \mathbb{N} \setminus \left\{ 1 \right\}[/itex]. Then [itex]A[/itex] and [itex]B[/itex] are both infinite sets, and they share all elements except for 1. Or is this not what you mean by "differ"?
     
  4. Oct 3, 2008 #3

    CRGreathouse

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    I take your question as:

    "Given two infinite sets A and B, is the symmetric difference [itex](A\cup B)\setminus(A\cap B)[/itex] ever nonempty?"

    in which case the above post gives an example of when this can occur.
     
  5. Oct 3, 2008 #4
    Please forgive my ignorance, but what does \ mean in this context?
     
  6. Oct 3, 2008 #5
    Let' say A = {0, 1, 2, ...} and B = {1, 2, 3, ...}. A and B are both infinite, but A and B "differ" by one element. Namely 0.
     
  7. Oct 3, 2008 #6

    CRGreathouse

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    Set subtraction (thus the LaTeX command \setminus). {1, 2, 3} \ {2} = {1, 3}.
     
  8. Oct 3, 2008 #7

    HallsofIvy

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    It's the "set difference". A\B is "All values that are in A but not in B". Think "A take away any members of A intersect B".
     
  9. Oct 3, 2008 #8
    Thanks, all.
     
  10. Oct 8, 2008 #9
    There isn't an inverse function to \, is there?
     
  11. Oct 8, 2008 #10
    Yes there is, the union. A \ B U B = A.
     
  12. Oct 26, 2008 #11
    Not quite; only if B is contained in A.
     
  13. Oct 26, 2008 #12

    HallsofIvy

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    Only if B is a subset of A.

    Using "is contained in" to mean "is a subset of" can cause an awful lot of confusion.
     
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