# Proof verification : limit of the sequence.

im taking my first course in real analysis this summer and I would like some feedback on proof writing. Thank you.

Let $\{x_n\}_n$ be a sequence of real number such that $x_n>0$ for all $n \in \textbf{}N$ and $\lim_{x\to\infty}x_n=l>0.$

1) Let $r<l$. Show that there exist an $N(\epsilon) \in \textbf{N}$ such that: $$n \geq N(\epsilon) \implies x_n > r$$ .

2) Show that $\inf \{ x_n \ | \ n \in \textbf{N} \} > 0$

$\textbf{Question 1.}$

We have: $$\lim_{x\to\infty}x_n=l>0.$$

The definition of the limit say that : $$\forall \epsilon>0, \ \ \ \exists N(\epsilon) \in \textbf{N} : |x_n - l|<\epsilon \ , \ \ \ \forall n>N(\epsilon).$$

Let $\epsilon = l - r$ and then : $$|x_n - l |< \epsilon.$$ $$\iff -\epsilon < x_n - l< \epsilon \ \ , \ \ \ \ \forall n>N(\epsilon).$$ $$\iff l-\epsilon<x_n<l+\epsilon \ \ , \ \ \ \ \ \forall n>N(\epsilon).$$ $$\iff l-(l - r)<x_n<l+\epsilon \ \ , \ \ \ \ \ \forall n>N(\epsilon).$$ $$\iff r<x_n<l+\epsilon \ \ , \ \ \ \ \ \ \ \ \ \ \ \ \forall n>N(\epsilon).$$

$\textbf{Question 2.}$ I'm not sure if I understand the question correctly.

Suppose that $$\inf\{x_n \ | \ n \in \textbf{N} \} \leq 0$$ and $${x_n}>0 \ \forall n \in \textbf{N}$$

We conclude that $$\inf\{x_n \ | \ n \in \textbf{N} \} \geq 0.$$

Any feedback is appreciated. Thank you very much.

• Question 1. Your proof in correct. Just to give you another tool. You could also invoke the triangle inequalities |x-l|>|x|-l. Question 2. You have not shown that $x_n$ does not get arbitrarily close to 0. In which case inf {$x_n$} = 0. Since all n>N are close to l. there are only finely many n that could be close to 0. Which means that there is some minimum distance $x_n$ can get to 0 before it starts converging on l. Jun 8, 2016 at 15:27
• So the fact $x_n > 0$ for all $n$ is not enough to show $\inf |{x_n : n \in \mathbb{N} \} > 0$, only $\geq 0$. Take for example the sequence $x_n = 1/n$. To show it isn't $0$, you'll have to appeal to $x_n > 0$ and $\ell > 0$. HINT: Suppose for contradiction $\inf |{x_n : n \in \mathbb{N} \} = 0$. What does that mean (you want a phrasing in terms of $\epsilon, n$)?
– AJY
Jun 8, 2016 at 15:27
• Thanks guys, I see my confusion now in question 2. Jun 8, 2016 at 16:17

(1) is fine — good work.

With (2), somehow you're confused, though I'm not sure how/why. The two conditions you state in your 'proof' aren't necessarily a contradiction at all. Example: if $y_n = \frac 1 {(n+1)}$, then $\inf_n y_n = 0$ and all $y_n > 0$. However, this sequence converges to $0$. So to prove (2) you'll need to use what's different about $(x_n)_n$.

You can use what you just proved in (1). Take $r = l/2$, so $l > r > 0$. By (1) there is $N$ such that for all $n > N$, $x_n \ge r$. Then $$\inf_n x_n = \min(\,\min_{i \le N} x_i, \inf_{n > N} x_n).$$ Because $x_n \ge r$ for every $n > N$, it follows that $\inf_{n > N} x_n \ge r$; thus, $$\inf_n x_n \ge \min(\,\min_{i \le N} x_i, r). \tag{*}$$ But all of the finitely many $x_i, i \le N$, are greater than $0$, so their min is also greater than $0$. Hence the righthand side of (*) is greater than $0$.

The proof of the first proposition is correct the proof of the second however is not. For two why not proceed as follows.

For convinience i denote $$\inf \{x_n:n\in\mathbf{N}\}$$ by $$\beta$$.

Proof. Now choose any $$\epsilon>0$$, evidently $$\beta+\epsilon$$ cannot be an lower bound for $$\{x_n:n\in\mathbf{N}\}$$ as $$\beta$$ be definition is the least upper bound. So $$x_r<\beta+\epsilon$$, for some $$r\in\mathbf{N}$$, but $$0 thus $$0<\beta+\epsilon$$. Since our choice of $$\epsilon$$ was arbitrary it follows that $$\beta+\epsilon>0,\forall \epsilon>0$$, implying $$\beta>0$$.

$$\blacksquare$$

Have Fun with Analysis !

Your proof of (1) is good.

Take $0<\varepsilon<l$; your proof of (1) tells you that there exists $m$ such that, for $n>m$, $x_n>\varepsilon$. Thus $$\inf\{x_n:n\in\mathbb{N}\}\ge \min\{\varepsilon,x_0,x_1,\dots,x_n\}>0$$ You just have to fill in some detail.