# Problem in understanding upper limit and lower limit in Real Analysis  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 $$u_{r_{n}} for all $$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$$"??

• 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 $r_2$. – User203940 Sep 16 at 21:19
• 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 $u_{r_n} \leq u^* < u^* + 1/n$. – User203940 Sep 16 at 21:21
• @User203940, how is $u_{r_{n}}\leq u^*$? – MrAP Sep 16 at 21:34
• You're correct this is not clear. Give me a moment to fix my answer – User203940 Sep 16 at 21:40
• I've hopefully fixed my answer now @MrAP – 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.

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.

• 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/… – User203940 Sep 16 at 21:57