I have been working on and learning about subsequential limits but I still have some problems that I don't know how to show: * Every bounded sequence has at least one subsequential limit * Non-bounded sequence may not have a subsequential limit in real numbers * If some element b is in a sequence infinitely, then b is the subsequential limit of the sequence Can someone help me tackle these problems?
Asked
Active
Viewed 223 times
0
-
If these are sequences of real numbers, you could first show that every sequence has a monotone subsequence, then use the fact that a monotone bounded sequence converges. – Sep 26 '17 at 06:38
1 Answers
0
Hints:
$(1)$$\;\;$For an unbounded sequence, consider the simplest one: $$1,2,3,...$$ Does any infinite subsequence have a limit?
$(2)$$\;\;$If some value $b$ occurs infinitely often, just take the subsequence of terms whose value is $b$. What does that subsequence look like?
$(3)$$\;\;$For the case of a bounded infinite sequence, consider two cases . . .
First suppose the set of values of the terms is a finite set. Argue that some value must occur infinitely often.
Next suppose the set of values of the terms is an infinite set. For this case, apply the theorem that every bounded infinite set has a limit point. Then argue that there must be an infinite subsequence converging to that limit point.
quasi
- 58,772
-
(1)So if I take the sequence to be the set of natural numbers which is unbounded above and take the subsequence of that(for example all odd numbers). This doesn't have a limit amongst real numbers and so is a correct example of the problem? – user442291 Sep 26 '17 at 11:34
-
@user442291: No, you need to show that all infinite subsequences of the sequence $$1,2,3,...$$ fail to converge. – quasi Sep 26 '17 at 11:36
-
Ok I got that one, now for (2). I understand that I take from a sequence such a subsequence where all values are b, so let's say $x_{n_k}={b,b,...}$. So the limit of that is also b and therefore the subsequential limit is b. – user442291 Sep 26 '17 at 13:16
-