Proof that a sequence has two subsequential limits

[Mr Davis 97]

Suppose I have the sequence $a_n = 2^{(-1)^n}$. So $\displaystyle (a_n) = (\frac{1}{2},2,\frac{1}{2},2,\frac{1}{2},2,\frac{1}{2},2,...)$. Clearly, this sequence has two subsequential limits, $\displaystyle \{\frac{1}{2},2 \}$. This clear from observation, but I'm not sure how I can be sure and show that these two are subsequantial limits and the only subsequential limits.

 

[andrewkirk]

To show that they are subsequential limits just choose the two obvious subsequences that have them as limits because they are constant on that value.

To show there are no other subseq limits, let x be a real number that is neither 1/2 nor 2 and show that there is some $\epsilon>0$ such that every member of the whole sequence differs from x by at least $\epsilon$. It follows that every member of any subsequence also differs from x by at least $\epsilon$.

 

[Mr Davis 97]

To show that they are subsequential limits just choose the two obvious subsequences that have them as limits because they are constant on that value.

To show there are no other subseq limits, let x be a real number that is neither 1/2 nor 2 and show that there is some $\epsilon>0$ such that every member of the whole sequence differs from x by at least $\epsilon$. It follows that every member of any subsequence also differs from x by at least $\epsilon$.

Would this work for sequences in general? For example, if I give you a sequence $(a_n)$ and the set of subsequential limits (that I find perhaps by inspection), is there a general way to prove that there are no other subequential limits, and that the list is exhaustive?

 

[andrewkirk]

The reason the above works is that the set of values of sequence elements has finite cardinality (cardinality is 2). The same approach can be taken for any sequence where that set of values has finite cardinality. It won't work where the set is infinite.

 

[Mr Davis 97]

The reason the above works is that the set of values of sequence elements has finite cardinality (cardinality is 2). The same approach can be taken for any sequence where that set of values has finite cardinality. It won't work where the set is infinite.

Is there an example of sequence whose set of values of sequence elements has infinite cardinaliity, but a finite number of subsequential limits, thus needing to prove that set of subsequential limits is exhaustive?

For example, what about $a_n = 2^{(-1)^n(2n+1)}$?

 

[andrewkirk]

Many sequences will be of that type.
For instance, any monotonic sequence that asymptotically approaches a limit, such as $a_n=1/n$.

 

[Mr Davis 97]

Many sequences will be of that type.
For instance, any monotonic sequence that asymptotically approaches a limit, such as $a_n=1/n$.

But suppose that a sequence has infinitely many distict sequence elements, and that is has, say, two elements in its set of subsequential limits. How could you verify that there are no more than 2 subsequential limits?

 

[andrewkirk]

But suppose that a sequence has infinitely many distict sequence elements, and that is has, say, two elements in its set of subsequential limits. How could you verify that there are no more than 2 subsequential limits?

I doubt there is any general method for doing that. But in any particular case it may be possible to produce a case-specific proof based on how the sequence is specified. My guess is that in most cases it may be reasonably straightforward to produce a proof, but that the nature of the proofs would differ between cases.

Take for instance the sequence whose nth element is 1 if n is odd and 1/n if n is even. I imagine you would find it fairly easy to show the only subsequential limits of this are {0,1} despite there being infinitely many values.

 

[mathwonk]

a limit is a number your sequence elements get (and stay) arbitrarily close to. doesn't that pretty much do it? i.e. you cannot get closer than 1/4 say to anything but 1/2 and 2 with that sequence, but you can easily get and stay as close as you like to those two.