# Sub-sequential limit point

• Jun 20th 2010, 02:42 AM
math.dj
Sub-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 AM
Opalg
Quote:

Originally Posted by math.dj
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?

Yes (to both questions). Use the fact that R is separable. A closed subset of R has a countable dense subset. Construct a sequence which visits each element of that subset infinitely often.
• Jun 20th 2010, 05:02 AM
HallsofIvy
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: $a_n= \frac{1}{n}$ if n is odd, $a_n= \frac{n-1}{n}$ if n is even.
• Jun 21st 2010, 08:18 AM
Idealconvergence
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 AM
math.dj
are you trying to say there is a bijection between N and Q intersection [0,1]?
• Jun 25th 2010, 08:19 PM
Idealconvergence
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 AM
HallsofIvy
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, $g\circ f^{-1}$ is a bijection from A to B.