Proof verification : limit of the sequence.

Question

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.

Proof by contradiction:

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

Clearly we have an contradiction.

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

---

Any feedback is appreciated. Thank you very much.

Answer 1

(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 $0$.

Answer 2

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 $0<x_r$ 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 !

Answer 3

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.