Proving Convergence of Real Number Sequences with Metric Equations

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muzak
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Homework Statement


Prove that lim[itex]_{n} p_{n}= p[/itex] iff the sequence of real numbers {d{p,p[itex]_{n}[/itex]}} satisfies lim[itex]_{n}[/itex]d(p,p[itex]_{n}[/itex])=0

Homework Equations





The Attempt at a Solution


I think I can get the first implication. If [itex]lim_{n} p_{n}[/itex]= p, then we know that d(p,p[itex]_{n}[/itex]) = d(p[itex]_{n}[/itex],p) <[itex]\epsilon[/itex]. Then given [itex]\epsilon[/itex] > 0 and some N, for n>N we have |d{p,p[itex]_{n}[/itex]-0|<d{p,p[itex]_{n} = d(p_{n},p) < \epsilon[/itex].

I'm having a little trouble with the backwards implication, do I just do what I did up above but backwards sorta? Or should I pick some p[itex]_{n}[/itex] and show that it converges to 0, like 1/n or something.
 
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You are very much on the right track, but your "proofs" are still quite sloppy and it's hard to see whether you're reasoning circularly here.
For example,
I think I can get the first implication. If [itex]lim_{n} p_{n}[/itex]= p, then we know that d(p,p[itex]_{n}[/itex]) = d(p[itex]_{n}[/itex],p) <[itex]\epsilon[/itex].
Actually, you know that for any [itex]\epsilon > 0[/itex] there is an N such that this is true for all n > N.

Maybe it helps if you first write out exactly what you need to prove, in the form:

For all [itex]\epsilon > 0[/itex], I need to prove that ...(if there exists an N such that ... there also exists an N' such that ...)... , and that (the other implication).