Build a sequence $x_n$ such that $PL\{x_n\}=[0,1]$. $PL$ of a sequence is the set of partial limits...
I tried using the harmonic series but it wouldn't work. Any suggestion?
Asked
Active
Viewed 139 times
1
-
what is $PL \lbrace x_n \rbrace $ ? – rafalpw Mar 01 '15 at 10:25
-
I edited it.... – Meitar Mar 01 '15 at 10:27
-
Is a "partial limit" the limit of a subsequence? Is each $x_n$ a real number? By $PL{x_n}$ do you mean the set of partial limits of the sequence? Your question still is not clear. – Rory Daulton Mar 01 '15 at 11:27
2 Answers
2
Try $ \mathbb{Q} \cap [0,1]$.
rafalpw
- 600
-
-
-
I mean taking $\lbrace x_n \rbrace = \mathbb{Q} \cap [0,1] $ , because we know, that this set is countable (can be arranged in a sequence). – rafalpw Mar 01 '15 at 11:00
1
This answer assumes that a "partial limit" is the limit of a subsequence, each $x_n$ is a real number, and $PL\{x_n\}$ is the set of partial limits of the sequence.
One possible sequence is $$1,$$ $$\frac 12,1,$$ $$\frac 14,\frac 12,\frac 34,1,$$ $$\frac 18,\frac 14,\frac 38,\frac 12,\frac 58,\frac 34,\frac 78,1,$$ $$\ldots,$$ $$\frac 1{2^k},\frac 2{2^k},\frac 3{2^k},\ldots,\frac {2^k-1}{2^k},1,$$ $$\ldots$$
It's not very hard to come up with a formula for $x_n$: I'll leave it to you.
There are many variations on this, of course. One is to use powers of ten rather than powers of two:
$$1,$$ $$0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1.0,$$ $$0.01,0.02.0.03,\ldots,0.99,1.00,\ldots$$
It should be clear why every real number between zero and one is a limit of a subsequence of either of these sequences.
Let me know if you need more detail, or if you need a sequence where all the $x_n$ are distinct.
Rory Daulton
- 32,288
-
$$x_0=0, \qquad x_{q(q-1)/2+p}=\frac{p}{q} \qquad \text{ for } 1\leq p \leq q\text{ and } q=1,2,\ldots$$ – Surb Mar 17 '15 at 21:08