Sequences and subsequences, Part IV

Recall from part III that we are using the following notation, which may not be entirely standard:

Let $X$ be a non-empty set, and let $(x_n)_{n\in \N}$ be a sequence of elements of $X$. Given positive integers $n_1<n_2<n_3<\cdots$, define the infinite set $A \subseteq \N$ by $A=\{n_1,n_2,n_3,\dots\}$. Then $(x_n)_{n\in\A}$ denotes the subsequence $x_{n_1}, x_{n_2}, x_{n_3},\dots$ of $(x_n)$.

Note (again) that we are working with a one to one correspondence between infinite subsets of $\N$ and strictly increasing sequences of natural numbers here, and so every subsequence of $(x_n)$ does have this form. (However, with our conventions, if the sequence $(x_n)$ has repeats, then different subsets of $\N$ might give the same subsequence.) Note also that if $A=\N$ then (to our relief) the notation $(x_n)_{n\in A}$ is unambiguously equal to the original sequence $(x_n)_{n\in \N}$.

Now let's look at convergence of sequences of real numbers and connections with subsets of $\N$. (A lot of what we say below can be generalised to arbitrary metric spaces, replacing modulus by the relevant metric.)

Definition of convergence

Let $(x_n)_{n\in \N}$ be a sequence of real numbers, and let $L\in\R$. Then we say that the sequence $(x_n)$ converges to $L$ if, for all $\ve>0$, there exists $N\in\N$ such that, for all $n\geq N$, we have $|x_n-L|<\ve$.

It is standard that if $(x_n)$ converges to $L$, then so does every subsequence of $(x_n)$. I may revisit this fact below, and look at it in terms of subsets of $\N$. For now let us take this fact as standard. I will also omit a few details below, which I leave to the reader (for now). I may fill in a few of these details in follow-up posts. See if you can spot which details are missing!

In previous posts I have looked at various approaches to handling the "three-quantifier" definition of convergence above to try and make it more digestible to students. Here is another approach. 

Let $(x_n)_{n\in \N}$ be a sequence of real numbers, and let $L\in\R$. For each $\ve>0$, we can look at what I will temporarily call the "good" indices and the "bad" indices (depending on $\ve$). We set

$G_\ve = \{n\in\N: |x_n-L|<\ve\,\}$  and  $B_\ve=\{n\in\N: |x_n-L|\geq \ve\,\}.$

Let $\ve>0$. Then $\N$ is the disjoint union of the two sets $G_\ve$ and $B_\ve$. We do not know at first whether these sets are empty, non-empty, finite or infinite. However, at least one of them must be infinite, and moreover we can easily see that one of them is finite if and only if the other has a subset of the form $\{N,N+1,N+2,\dots\,\}$ for some $N \in \N$. This is because every non-empty finite subset of $\N$ has a maximum element, and we can take any $N$ strictly greater than the maximum element of that finite set. (If one of the sets is empty, then we can choose any $N\in \N$, because the other set is $\N$.)

Now let us revisit the definition of convergence to $L$. In view of the discussion above, we see that $(x_n)$ converges to $L$ if and only if, for all $\ve>0$, the set $B_\ve$ is finite.

Negating this definition, we see that $(x_n)$ does not converge to $L$ if and only if there exists an $\ve>0$ such that $B_\ve$ is infinite. (Of course this does not say whether $(x_n)$ converges at all, but if it does converge, then the limit is not $L$.) For such an $\ve$, we can set $A=B_\ve$ (an infinite subset of $\N$) and consider the subsequence $(x_n)_{n\in A}$. Then, every term of this subsequence is at least $\ve$ away from $L$.

This leads us to the following equivalent criterion for "non-convergence to $L$". The sequence $(x_n)$ fails to converge to $L$ if and only if $(x_n)$ has a subsequence $(y_n)$ such that the set $\{|y_n-L|:n\in\N\,\}$ is bounded below by some positive constant. [Have you spotted an omitted detail here?]

This last condition on the subsequence $(y_n)$ is equivalent to saying that $\inf \,\{|y_n-L|:n\in\N\,\} > 0,$ i.e., there exists $\ve>0$ such that, for all $n\in\N$, we have $|y_n-L|\geq \ve$. 

This "bad" property of the subsequence $(y_n)$ is much stronger than just saying that $(y_n)$ fails to converge to $L$. Because every subsequence $(z_n)$ of $(y_n)$ has the same property: indeed, for any such subsequence $(z_n)$ of $(y_n)$ we will have

$\inf \{|z_n-L|:n\in\N\,\}  \geq \inf \{|y_n-L|:n\in\N\,\} > 0.$

But, in particular, no subsequence of $(y_n)$ can converge to $L$.

We now see that, if $(x_n)$ does not converge to $L$, then $(x_n)$ has a subsequence $(y_n)$ such that no subsequence of $(y_n)$ converges to $L$.

However, if $(x_n)$ does converge to $L$, then every subsequence $(y_n)$ of $(x_n)$ also converges to $L$. Thus each such $(y_n)$ has a subsequence which converges to $L$ (indeed $(y_n)$ is a subsequence of itself, and that is enough to prove existence.)

We have now recovered the standard result that $(x_n)$ converges to $L$ if and only if every subsequence of $(x_n)$ has at least one subsequence which converges to $L$.

For those who like cluster points (also known as limit points), $(x_n)$ converges to $L$ if and only if every subsequence of $(x_n)$ has $L$ as a cluster point.

More soon!