enter image description here

enter image description here

I am studying Real Analysis on my own. I am currently trying to understand sequences. I cannot understand the parts marked in blue in the second picture.

In the line: "Therefore there is a natural number $r_2>r_1$ such that $u_{r_{2}}>u^*-\frac{1}{2}$", how do we know $r_2>r_1$?

We know $u_{r_{n}}>u^*-\frac{1}{n}$ for all $n\in \mathbb{N}$. But how do we know that $u_{r_{n}}<u^*+\frac{1}{n}$ for all $n\in \mathbb{N}$ so that we can write $|u_{r_{n}}-u^*|<\frac{1}{n}$ in the line "$|u_{r_{n}}-u^*|<\frac{1}{n}<\epsilon$ for all $n\geq k$"??

  • $\begingroup$ Here's the idea: we know there are infinitely many n so that $u_n > u^* - 1$. Let $r_1$ be one of the $n$. Now we know there are infinitely many $n$ so that $u_n > u^* - 1/2$. Since there are infinitely many, there has to be one greater than $r_1$. Choose that one for $r_2$. $\endgroup$ – User203940 Sep 16 at 21:19
  • $\begingroup$ For the next part, we know $u^* - 1/n < u_{r_n} \leq u^*$. Since $1/n > 0$, we can add it to $u^*$, so $u_{r_n} \leq u^* < u^* + 1/n$. $\endgroup$ – User203940 Sep 16 at 21:21
  • $\begingroup$ @User203940, how is $u_{r_{n}}\leq u^*$? $\endgroup$ – MrAP Sep 16 at 21:34
  • $\begingroup$ You're correct this is not clear. Give me a moment to fix my answer $\endgroup$ – User203940 Sep 16 at 21:40
  • $\begingroup$ I've hopefully fixed my answer now @MrAP $\endgroup$ – User203940 Sep 16 at 21:50

I just realized I ended up answering the question in the comments, so here's maybe a more fleshed out version of what I wrote in the comments.

Let's suppose $u^*$ is the least upper bound. If we subtract $1$ from it, then it is no longer an upper bound, meaning there is subsequential limit $\ell_1$ (meaning there is a subsequence which converges to $\ell_1$) with the property that $\ell_1 > u^* - 1$. Since this is a subsequential limit, we can use the definition to figure out there has to be infinitely many $u_n$ which satisfy the condition that $u_n > u^* - 1$. Let's choose $r_1$ to be one of these.

Now let's try subtracting $1/2$, so we look at $u^* - 1/2$. Again, we know that there is a subsequential limit $\ell_2$ so that $\ell_2 > u^* - 1/2$, so there are infinitely many $n$ which satisfy the condition that $u_n > u^* - 1/2$. Now let's suppose they are all less than or equal to $r_1$. Well that means that we have infinitely many $n$ between $1$ and $r_1$, but this is silly because we know there are only finitely many. So there must actually be infinitely many $n$ which satisfy the condition that $n > r_1$ and $u_n > u^* - 1/2$. Choose one of these and set $r_2$ to be this.

Notice there's nothing special about these two cases. I could repeat this same kind of argument for $u^* - 1/3$ and find an $r_3 > r_2$ so that $u_{r_3} > u^* - 1/3$, and then do it for $1/4$, $1/5$, so on and so forth. So we have an increasing sequence $r_1 < r_2 < \cdots < r_n < \cdots $ which satisfies the condition that $u_{r_n} > u^* - 1/n$.

EDIT: I was not fully paying attention when I wrote the answer. It is a good question to ask why $u_{r_n} < u^* + 1/n$. I will edit this with an answer later.

EDIT 2: Here's an updated answer.


$$ S = \{\ell \in \mathbb{R} : \exists \{n_j\}_{j=1}^\infty \text{ such that } u_{n_j} \rightarrow \ell\}.$$

We take the least upper bound of this set, so $u^*$ is the least upper bound of $S$. Let's go back to our choices of $\{r_n\}_{n=1}^\infty$ and let's be more careful about them. Notice that we have

$$u^* - 1 < \ell_1 \leq u^* < u^* + 1.$$

By definition of convergence, there are infinitely many $u_n$ which satisfy

$$ u^* - 1 < u_n < u^* + 1.$$

Choose $r_1$ to be an $n$ which satisfies this. Now if we subtract $1/2$, there is an $\ell_2$ so that

$$ u^* - 1/2 < \ell_2 \leq u^* < u^* + 1/2.$$

Again, there are infinitely many $u_n$ which lie between these bounds by the definition of convergence. Choose $r_2 > r_1$ so that this holds. Now repeat the process to create an increasing sequence $\{r_n\}_{n=1}^\infty$ so that

$$ |u_{r_n} - u^*| < 1/n.$$

Use the Archemedean principle to get convergence.

| cite | improve this answer | |
  • $\begingroup$ Good pointing out that we may not necessarily have $u_{r_n} \leq u^*$. I'm not sure if the argument in the textbook is correct as is -- I think you have to make sure your choices of $u_{r_n}$ lies in the appropriate open interval. See for example math.stackexchange.com/questions/581128/… $\endgroup$ – User203940 Sep 16 at 21:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.