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Continuity to the right.

  1. Nov 4, 2006 #1

    quasar987

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    My probability professor proved the property of "continuity to the right" of the repartition fucntion F by showing that F(x+1/n)--->F(x). But as I remember it, continuity to the right means that for any sequence {a_n} of elements of (x,+infty) that converges to x, F(a_n) converges to F(x). Is there some subtlety I'm not aware of by which showing F(x+1/n)--->F(x) is sufficient to show it works for any sequence?
     
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  3. Nov 4, 2006 #2

    arildno

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    Well, you are right.
    If your professor's proof is correct, then his proof cannot essentially rely on other properties of the 1/n-sequence than that it converges to zero; i.e, that is, it is only the DISTANCE a given point x+1/n has from x that is relevant in the actual proof.

    It might be that it is easiest to use 1/n as an EXAMPLE of a sequence, but the proof shouldn't crucially rest upon that particular choice.
     
    Last edited: Nov 4, 2006
  4. Nov 6, 2006 #3

    quasar987

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    He agreed (by email) that it is insufficient to prove it for 1/n in the case of any function, but because the partition function F(x) is an increasing function, it is sufficient. Do you see why?
     
  5. Nov 6, 2006 #4

    arildno

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    Okay, note that since F is INCREASING, it means that the function values for any points squeezed in between 1/n and 1/(n+1) must be squeezed in between the function values F(1/n) and F(1/(n+1)).
    Thus, any epsdilon/delta squeeze valid for the 1/n sequence must be valid for any other sequence converging to 0 as well, since its terms are necessarily squeezed in between the harmonic sequence terms.
     
  6. Nov 6, 2006 #5

    quasar987

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    I see it thanks.
     
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