Any bounded sequence of real numbers contains at least one accumulation point. If it doesn't converge it has more than one. In fact, $$a_n \equiv n (mod m)$$ has exactly m limit points. Question: Can a bounded sequence of real numbers have infinitely many limit points?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ What is your definition of a limit point? $\endgroup$– zhw.Commented Aug 12, 2017 at 0:28
-
$\begingroup$ Limit points of its image. $\endgroup$– Hyobin LeeCommented Aug 12, 2017 at 0:59
Add a comment
|
5 Answers
$\begingroup$
$\endgroup$
2
1, 1,1/2, 1,1/2,1/3, 1,1/2,1/3,1/4, ...
-
$\begingroup$ What sequence is this? What are its infinitely many limit points? $\endgroup$ Commented Aug 11, 2017 at 15:26
-
$\begingroup$ @Benjamin $1,1/2,1/3,\dots $. (It is obviously not an injective sequence.) $\endgroup$ Commented Aug 11, 2017 at 22:02
$\begingroup$
$\endgroup$
2
Take an infinite family of bounded sequences in $[0, 1]$ converging to $1/k$ for each $k \in \mathbb{N}$.
Interweave the sequences (dovetail them) to produce an explicit example.
That is, for $a_k \rightarrow 1$, $b_k \rightarrow 1/2$, $c_k \rightarrow 1/3$, $\ldots$:
$a_1, a_2, a_3, a_4 \ldots$
$b_1, b_2, b_3, b_4 \ldots$
$c_1, c_2, c_3, c_4, \ldots$
$\vdots$
becomes:
$a_1, b_1, a_2, c_1, b_2, a_3, \ldots$
where this final sequence has the desired property.
answered Aug 11, 2017 at 12:34
-
1$\begingroup$ Now I recall that method from set theory. Thank you! $\endgroup$ Commented Aug 11, 2017 at 12:36
-
$\begingroup$ @HyobinLee Yes, it is often encountered in showing $\mathbb{Q}$ (or, first, $\mathbb{Q}^{+}$) is countable. As a reminder of what this looks like in practice, I added a few additional details. Hope this helps! $\endgroup$ Commented Aug 11, 2017 at 12:37
$\begingroup$
$\endgroup$
1
The sequence
$$0,1,0,1/2,1,0,1/3,2/3,1, 0,1/4,2/4,3/4,1, \dots$$
has every point in $[0,1]$ as an accumulation point.
-
$\begingroup$ Seems like this could be the easiest way to explain the answer to a beginner. Thanks! $\endgroup$ Commented Aug 12, 2017 at 0:23
$\begingroup$
$\endgroup$
4
Yes. Take the sequence that enumerate all rational between $[-1,1]$.
A possible sequence that enumerate rational in $[0,1]$: $$x_{\frac{q(q-1)}{2}+p}=\frac{p}{q}.$$
-
$\begingroup$ That was quick! Can you show me an explicit construction? $\endgroup$ Commented Aug 11, 2017 at 12:30
-
$\begingroup$ Thanks! I'll have to think how to show that the indices don't overlap though... Btw I came across this gem in a related thread. en.m.wikipedia.org/wiki/Calkin–Wilf_tree $\endgroup$ Commented Aug 11, 2017 at 12:48
-
$\begingroup$ @HyobinLee : your link looks wrong. $\endgroup$– SurbCommented Aug 11, 2017 at 12:51
-
$\begingroup$ Yeah it broke. You should copy and paste the whole adress including the -Wilf_tree part! $\endgroup$ Commented Aug 11, 2017 at 12:53
$\begingroup$
$\endgroup$
2
There are uncountably many ways to do this. One way, which does not enumerate all of $\Bbb Q$, is to write the natural numbers in their usual decimal form $0,1,...,9,10,11,12,..., 6792,6793,...,$ and assign to each one the reversed decimal fraction, thus generating the sequence of decimal rational numbers $0.0,0.1,...,0.9,0.01,0.11,0.21,..., 0.2976,0.3976,...$. It is easily seen that every real number in the interval $[0\,\pmb,\,1]$ is distant by no more than $10^{-n}$ from one of the numbers in the first $10^n$ terms of this sequence.
answered Aug 11, 2017 at 21:37
-
$\begingroup$ (Which amounts to enumerating a dense set of rationals.) $\endgroup$ Commented Aug 11, 2017 at 22:00
-
$\begingroup$ I must have learned about this in set theory 10 years ago and forgotten about it. What a clever trick. Thank you! $\endgroup$ Commented Aug 12, 2017 at 0:21