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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 $n\geq k$"??

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  • $\begingroup$ 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
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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.

Consider

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

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.

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  • $\begingroup$ 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

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