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Given a sequence: $$ \begin{cases} \{x_n\} = \left\{1, \frac{1}{10}, \frac{2}{10},\cdots,\frac{9}{10}, \frac{1}{10^2}, \frac{2}{10^2},\cdots,\frac{99}{10^2}, \cdots, \frac{1}{10^n}, \frac{2}{10^n}\cdots \frac{10^n-1}{10^n}, \cdots\right\} \\ n \in \Bbb N \end{cases} $$ Find the subsequential limits of $\{x_n\}$. Find $\lim\sup\{x_n\}$ and $\lim\inf\{x_n\}$.

For the second part it seems obvious since for the $\lim\inf$ just take a subsequence: $$ x_{n_k} = \frac{1}{10}, \frac{1}{10^2}, \cdots, \frac{1}{10^n}, \cdots \\ \lim_{n_k \to\infty} x_{n_k} = \lim_{n\to\infty}\inf x_n = 0 $$

To find limit supremum one may observer that: $$ \lim_{n_p \to \infty} x_{n_p} = \left\{1, \frac{9}{10}, \frac{99}{100}, \cdots, \frac{10^n-1}{10^n},\cdots\right\} =\lim_{n\to\infty}\sup x_n = 1 $$

But for the first part it's not obvious at a first glance. I believe there is a way to rearrange that sequence to find the set of subsequential limits.

The limit is not affected by cutting off a finite number of the terms from the sequence, and thus the subsequences should not be affected as well. And hence there is infinitely many starting points for the subsequences.

Based on that it feels like that set of subsequential limits will form a set of rational numbers in the range of $[0, 1]$ but I'm not quite sure.

How do I find the set of subsequential limits for $x_n$?

user376343
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roman
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2 Answers2

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Find the set of partial limits.

First of all: As English is not my native language, I might misunderstand this question.

If I understand the term "partial limit" correctly, it is $\lim\limits_{k\to\infty}x_{n_k}$ while $n_k$ is some strictly strictly monotonously rising sequence.

If my understanding is correct, the set of partial limits is $[0,1]$:

Using the sub-sequence $1,\frac{1}{10}\dots\frac{1}{10^n}\dots$ you can show that 0 is a partial limit.

And using the sub-sequence $\frac{9}{10},\frac{99}{100}\dots\frac{10^n-1}{10^n}\dots$ you can show that 1 is a partial limit.

You may also take any real number in the range $]0,1[$ and approximate the number using decimal digits:

Example: $\frac{\sqrt{2}}{2}=0.70710678\dots$

Now take the following sub-sequence: $0.7,0.70,0.707,0.7071\dots$

... which can be written as: $\frac{7}{10},\frac{70}{100},\frac{707}{1000},\frac{7071}{10000},\dots$

To prove that this sub-sequence really has the limit $\frac{\sqrt{2}}{2}$ you have to show that for each value $\epsilon>0$ some value $k_0$ exists so that $|\frac{\sqrt{2}}{2}-x_{n_k}|<\epsilon$ for all $k>k_0$.

Choose any value $k_0$ with $\frac{1}{10^{k_0}}<\epsilon\iff k_0>-log_{10}\epsilon$.

Martin Rosenau
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  • Thank you for the answer, that clarifies a lot. I know it's a bad practice to change the question after an answer received, but I would like to change "a set of partial limits" to "the set of partial limits" which would mean all partial limits if you don't mind. As for the partial limit you did understand it correctly. – roman Dec 10 '18 at 20:01
  • ok, someone has already done it – roman Dec 10 '18 at 20:02
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    These are often called "subsequential limits," I believe. – saulspatz Dec 10 '18 at 20:07
  • @saulspatz, you are right, thanks for suggesting the right term. – roman Dec 10 '18 at 20:11
  • @saulspatz I've found some mathematics page in the internet that says that "partial limit" is also a valid term for this. – Martin Rosenau Dec 11 '18 at 06:32
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For any $r \in [0, 1)$, let $r_n = \dfrac{\lfloor 10^nr \rfloor}{10^n} $.

Then $10^nr-1 \le \lfloor 10^nr \rfloor \le 10^nr$ so that $r-10^{-n} \le r_n \le r$ and $r_n$ is in your sequence.

Therefore $r$ is the limit of this subsequence.

Also, $\dfrac{10^n-1}{10^n} \to 1$ so $1$ is also a limit.

Therefore $[0, 1]$ is the limit set.

marty cohen
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