Does Showing F(x+1/n)--->F(x) Prove Continuity to the Right for Any Sequence?

[quasar987]

My probability professor proved the property of "continuity to the right" of the repartition function 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?

 

[arildno]

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.

 

[quasar987]

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?

 

[arildno]

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.

 

[quasar987]

I see it thanks.