## Sequence which has countably many convergent subsequences.

### August 22, 2011

This post is about a question I just thought about that I thought had a pretty neat solution.  The values we consider below are all in the Reals.

So, we’re all familiar with sequences which have subsequences which converge to two different values; for example, take $\{1,-1,1,-1,1,-1,\dots\}$ which has a subsequence converging to 1 and a subsequence converging to -1.  Similarly, we can construct a sequence containing subsequences converging to three different values in a similar way; for example: $\{-1,0,1,-1,0,1, \dots\}$.  Indeed, for any finite number, we can make a sequence containing subsequences converging to that number of different values.

The obvious next question is: can a sequence have subsequences which converge to a countable number of different values?  What about an uncountable number of different values?

For the former question, I thought of this solution.  First, enumerate your countable list of distinct values (think of the integers or the rationals, for example) as $\{q_{1}, q_{2}, \dots\}$.  To construct the sequence, we do the following:

• Let $a_{1} = 0$.
• Let $a_{b} = 0$ if $b$ is not some positive power of a prime (for example, if $b$ is 15, 21, 35, 100, etc.)
• For $a_{k}$ where $k$ is a positive power of a prime, we do the following: If $k = p_{i}^{r}$ for $r\in {\mathbb N}$ and $p_{i}$ is the $i$-th prime (so $p_{1} = 2$, $p_{2} = 3$, $p_{3} = 5$, and so on) then set $a_{k} = q_{i}$ where $q_{i}$ is the $i$-th element from your list above.

How does this actually work?  Let’s take a simple example.  Let’s let our countable set be the set of natural numbers.  This is easily ordered as $\{1,2,3,4,5,\dots\}$ and so, for this example, $q_{1} = 1$, $q_{2} = 2$, $q_{3} = 3$, and so forth.  Our sequence is now:

$\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}, a_{8}, a_{9}, a_{10},\dots\}$

$= \{a_{1}, a_{2}, a_{3}, a_{2^{2}}, a_{5}, a_{2\cdot 3}, a_{7}, a_{2^{3}}, a_{3^{2}}, a_{2\cdot 5}, \dots\}$

$= \{0, q_{1}, q_{2}, q_{1}, q_{3}, 0, q_{4}, q_{1}, q_{2}, 0, \dots\}$

$= \{0, 1, 2, 1, 3, 0, 4, 1, 2, 0, \dots\}$

and if we kept going like this,

$= \{0, 1, 2, 1, 3, 0, 4, 1, 2, 0, 5, 0, 6, 0, 0, 1, 7, 0, 8, 0, 0, 0, 0, 0, 3, 0,\dots\}$

You kinda see the pattern going on here.  The 1 will appear infinitely many times, but the spaces between it grow exponentially.  Same for 2, 3, 4, and so on.  Thus, this sequence has the subsequence $\{n, n, n, n, \dots\}$ for any $n\in {\mathbb N}$ which obviously will converge to $n$.

Do you have another neat way to do this?  Encoding this in the primes was my first thought, but you never know!

As for uncountable, my guess is no.  Of course, thinking of the sequence as a SET, we would get the reals from the rationals which is uncountable, but as a SEQUENCE it is not so trivial I feel.  I don’t want to spend too much time thinking about this (as analysis is calling me!) but I’m sure the proof isn’t too crazy.

Edit: Of course, a nice example exists where a sequence has subsequences converging to an uncountable number of distinct point.  In fact, two nice examples exist, and both were provided to me by Brooke (as usual!).

Uncountable Example 1:

It was a bit difficult for me to see the first one, but after doing a nice little thought experiment everything became much clearer.  Here it is:

Let $\{q_{1}, q_{2}, \dots\}$ be an enumeration of the rationals.  The claim is that there is actually a subsequence which converges to any real number.  Think about it for a minute.

Here’s a sketch for the proof.  Take your favorite real number.  Let’s start easy and just say we want a subsequence which converges to $\pi$.  Let $a_{1}$ be the first element in our enumeration of the rationals above which is a distance less than 1 away from $\pi$; let’s say it’s $q_{k}$.  Now let $a_{2}$ be the first element in our enumeration which is after $q_{k}$ and which is a distance less than $\frac{1}{2}$ away from $\pi$

After that, you just keep picking the "next" element from the list that’s $\frac{1}{2^{n}}$ away from $\pi$.  There must be one due to the density of the rationals in the reals (or, assume not and see what happens!).  We end up with $\{a_{1}, a_{2}, \dots\}$ as a subsequence of our enumeration of the rationals which converges to $\pi$.  Obviously, replacing $\pi$ with any other real number works exactly the same.

Uncountable example 2:

This was one that Brooke presented to me that I liked quite a bit since it’s really easy to state.  Here’s the sequence (broken up by lines for clarity):

$0.0\,,\, 0.1\,,\, 0.2\,,\, \dots\,,\, 0.9,$

$0.00\,,\, 0.01\,,\, 0.02\,,\, \dots\,, \,0.09\,,\, 0.10\,,\, 0.11\,,\, \dots\,,\, 0.99\,,$

$0.000\,,\, 0.001\,,\, 0.002\,,\, \dots\,,\, 0.999\,, \dots$

This converges to any real number in $[0,1]$.  To see this is easy: decimal expand your chosen real number and pick the element "from each line" (as I’ve written it above) which is closest to it.  They’ll be at most $10^{-j}$ away (for some suitable $j$ depending on the line you’re on) and thus a subsequence exists which converges to whatever element you picked.