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Hi everyone after lost the fight with the exercise in my last question I decide to continue with the book; I hope someday answer it.

I finished some stack of questions and I appreciate is somebody could verify my answers, most of them are kinda easy and I feel as I repeated the definitions this is where I have troubles because I have the feeling of circularity (as usual) in my arguments.

Here are my attempts.

$(1)$ Let $(a_n)_{n=m}^{\infty}$ be a sequence of real numbers, such that $\,a_{n+1}>a_{n}$ for each $n$. Prove that whenever $n,m\in \mathbb{N}$ such that $m>n$ then we have $a_m>a_n$.

This exercise is specially obvious but sometimes in this I have more troubles with the argumentation.

Proof: Suppose $\,a_{k+1}>a_{k}$ for each $k,\,$ let $\, n < m$ where $n,m\in \mathbb{N}$ and we may assume by the sake of the contradiction that $a_m \le a_n$. We set $m= n+1$ which is possible since $\,n+1 \le m$. Then $\,a_{n+1} \le a_n$ contradicting our assumption. Hence $\,a_m > a_n$ for $\, m > n$, as desired.

$(2)$ Let $(a_n)_{n=m}^{\infty}$ be a sequence of real numbers, let $c$ be a real number and let $m' \ge m$ be an integer. Show that $(a_n)_{n=m}^{\infty}$ converges to $C \iff$ $(a_n)_{n=m'}^{\infty}$ converges to $c$.

Proof:

($\Rightarrow$) Suppose $(a_n)_{n=m}^{\infty}$ converges to $c$. Let $\,\epsilon>0\,$ be an arbitrary positive real number. Then there exists a $N\ge m$ such that $\,d(a_n,c)\le \epsilon$ for each $n \ge N$. We set $M:= max(N,m')$. Thus, the sequence $(a_n)_{n=m'}^{\infty}$ is eventually $\epsilon$-close to $c$; this is, for all $n \ge M$ we have $\,d(a_n,c)\le \epsilon$ where $M\ge m'$. Since $\epsilon>0$ was arbitrary it follows that $(a_n)_{n=m'}^{\infty}$ is eventually $\epsilon$-close to $c$ for each $\epsilon$, i.e., $(a_n)_{n=m'}^{\infty}$ converges to $c$ as desired.

($\Leftarrow$) Now suppose $(a_n)_{n=m'}^{\infty}$ converges to $c$. Let $\,\epsilon>0\,$ be an arbitrary positive real number. Then there exists a $N\ge m'$ such that $\,d(a_n,c)\le \epsilon$ for each $n \ge N$. Since $N \ge m'\ge m$, so $N\ge m$ and $\,d(a_n,c)\le \epsilon$. Hence the sequence $(a_n)_{n=m}^{\infty}$ is eventually $\epsilon$-close to $c$ to any arbitrary $\epsilon$.

$(3)$ Let $(a_n)_{n=m}^{\infty}$ be a sequence of real numbers, let c be a real number and let $k \ge 0$ be a non-negative integer. Show that $(a_n)_{n=m}^{\infty}$ converges to $C \iff$ $(a_{n+k})_{n=m}^{\infty}$ converges to $c$.

Proof: Since $(a_{n+k})_{n=m}^{\infty}$ is equivalent to $(a_{n})_{n=m'}^{\infty}$ where $m' = m+k$ the result follows from (2).

So, do you think is correct the manner as I proved that exercises. Thanks as usual.

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For problem (1), I would do it by induction on $m-n$.

Theorem: If $a_{n+1} > a_n$ for all $n$, then $a_{n+k} > a_n$ for all $n$ and for all $k \ge 1$.

Proof:

Since $a_{n+1} > a_n$, this is true for $k=1$.

Suppose it is true for some $k \ge 1$. We want to show that it is true for $k+1$.

This means showing that $a_{n+k+1} > a_n$ for all n.

By the induction hypothesis, $a_{n+k} > a_n$ for all $n$.

By the original statement about $a$, $a_{n+1} > a_n$ for all $n$. Putting $n+k$ for $n$, this means that $a_{n+k+1} > a_{n+k}$ for all $n$.

By transitivity of "$>$", $a_{n+k+1} > a_{n+k}$ and $a_{n+k} > a_n$ implies that $a_{n+k+1} > a_{n}$ for all $n$.

This finishes the proof.

To show that $a_m > a_n$ for all $m > n$, let $k= m-n$. Then $a_m = a_{m-n+n} = a_{(m-n)+n} = a_{k+n} > a_n$.

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