1) Is it Possible to find a sequence for which the set of sub-sequential limit points is [0,1]?

2)Is any closed set in R,the set of sub-sequential limit points of a sequence?

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- Jun 20th 2010, 02:42 AMmath.djSub-sequential limit point
1) Is it Possible to find a sequence for which the set of sub-sequential limit points is [0,1]?

2)Is any closed set in R,the set of sub-sequential limit points of a sequence? - Jun 20th 2010, 03:11 AMOpalg
- Jun 20th 2010, 05:02 AMHallsofIvy
In other words, for (a) construct a sequence that converges to 0 ({1/n} will do), a sequence that converges to 1 ({(n-1)/n} will do) and "blend" the two: $\displaystyle a_n= \frac{1}{n}$ if n is odd, $\displaystyle a_n= \frac{n-1}{n}$ if n is even.

- Jun 21st 2010, 08:18 AMIdealconvergence
a) Consider the sequence {x_n} such that there is a one to one correspondence between the elements of the sequence and the rational numbers in [0,1]. Then the sequence has sub-sequential limits [0,1].

b) Yes. Any finite set is closed. So try to find a sequence with finite no. of limit points. - Jun 25th 2010, 04:38 AMmath.dj
are you trying to say there is a bijection between N and Q intersection [0,1]?

- Jun 25th 2010, 08:19 PMIdealconvergence
Yes. Since the set of rational nos. Q are countable, we can arrange the rational nos Q∩[0,1] as the sequence {x1,x2,x3,.....}. Then for any no. α in [0,1] there is a sub-sequence from {xn} converging to α.

- Jun 26th 2010, 08:14 AMHallsofIvy
There exist a bijection between any two countable sets:

If A is countable then, by definition, there exist a bijection f:N-> A.

If B is countable then, by definition, there exist a bijection g:N-> B.

The function A->B, $\displaystyle g\circ f^{-1}$ is a bijection from A to B.