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rednalino
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Metric Space and Topology HW help!
Let X be a metric space and let (sn
)n be a sequence whose terms are in X. We say that (sn
)n converges to s [itex]\ni[/itex] X if
[itex]\forall[/itex] [itex]\epsilon[/itex] > 0 [itex]\exists[/itex] N [itex]\forall[/itex] n ≥ N : d(sn,s) < [itex]\epsilon[/itex]
For n ≥ 1, let jn = 2[(5^(n) - 5^(n-1))/4].
(Convince yourself that 5^(n) - 5^(n-1) is always divisible by
4, so the exponent in the definition is always a positive integer.) The first few terms of this
sequence are
2; 32; 33554432; 42535295865117307932921825928971026432
so you would reasonably expect this sequence to diverge with respect to the usual metric on
Q (the one given by the usual absolute value).
However, show that |j2n - (-1)|5 ≤ 5^(-n) where ||5 is the 5-adic absolute value.
My Attempt:
I started by writing the claim in terms of v5(j2n + 1). Then i tried to find a recurrence that looks like this:
(j2n+ 1)^5 = (2n+1+1) + (some other stuff).
I was thinking I can show that the sequence (jn)n is also Cauchy with respect to ||5, so in Q5,the completion of Q with respect to ||5, the sequence (jn)n converges to a number j [itex]\ni[/itex]Q5 such that j2 = -1. It follows that Q5 is not an ordered field, unlike the completion of Q with respect to the usual ||5, which is our old friend R.
Let X be a metric space and let (sn
)n be a sequence whose terms are in X. We say that (sn
)n converges to s [itex]\ni[/itex] X if
[itex]\forall[/itex] [itex]\epsilon[/itex] > 0 [itex]\exists[/itex] N [itex]\forall[/itex] n ≥ N : d(sn,s) < [itex]\epsilon[/itex]
For n ≥ 1, let jn = 2[(5^(n) - 5^(n-1))/4].
(Convince yourself that 5^(n) - 5^(n-1) is always divisible by
4, so the exponent in the definition is always a positive integer.) The first few terms of this
sequence are
2; 32; 33554432; 42535295865117307932921825928971026432
so you would reasonably expect this sequence to diverge with respect to the usual metric on
Q (the one given by the usual absolute value).
However, show that |j2n - (-1)|5 ≤ 5^(-n) where ||5 is the 5-adic absolute value.
My Attempt:
I started by writing the claim in terms of v5(j2n + 1). Then i tried to find a recurrence that looks like this:
(j2n+ 1)^5 = (2n+1+1) + (some other stuff).
I was thinking I can show that the sequence (jn)n is also Cauchy with respect to ||5, so in Q5,the completion of Q with respect to ||5, the sequence (jn)n converges to a number j [itex]\ni[/itex]Q5 such that j2 = -1. It follows that Q5 is not an ordered field, unlike the completion of Q with respect to the usual ||5, which is our old friend R.