Proof of the limit laws (Analysis)

Question

Hi everyone I'd like to know if my arguments for the next proof are sound or needs some changes to be correct. I hope they are not a little flaws.

Proposition (limit laws): Let $(a_n)_{n=m}^\infty$ and $\,(b_n)_{n=m}^\infty$ be convergent sequence of reals, and let $x$ and $y$ be the real numbers $x:= \lim_{n \rightarrow \infty}a_n$ and $\,y:= \lim_{n \rightarrow \infty}b_n$

a) The sequence $(a_n+ b_n)_{n=m}^\infty$ converges to $x+y$, i.e., $\lim_{n \rightarrow \infty}a_n+b_n=\lim_{n \rightarrow \infty}a_n+\lim_{n \rightarrow \infty}b_n$

b) The sequence $(a_n b_n)_{n=m}^\infty$ converges to $xy$, i.e., $\lim_{n \rightarrow \infty}a_nb_n=(\lim_{n \rightarrow \infty}a_n)(\lim_{n \rightarrow \infty}b_n)$

c) For any real number $c$, the sequence $(ca_n)_{n=m}^\infty$ converges to $cx$; in other words we have that: $\lim_{n \rightarrow \infty}ca_n=c\lim_{n \rightarrow \infty}a_n$

d) The sequence $(a_n- b_n)_{n=m}^\infty$ converges to $x-y$, i.e., $\lim_{n \rightarrow \infty}a_n-b_n=\lim_{n \rightarrow \infty}a_n-\lim_{n \rightarrow \infty}b_n$

e) Suppose that $y\not=0$ and that $b_n \not=0$ for all $n\ge m$. Then the sequence $(b_n^{-1})_{n=m}^\infty$ converges to $y^{-1}$; in other words: $\lim_{n \rightarrow \infty} b_n^{-1}=(\lim_{n \rightarrow \infty} b_n)^{-1}$

f) Suppose that $y\not=0$ and that $b_n \not=0$ for all $n\ge m$. Then the sequence $(a_n/b_n)_{n=m}^\infty$ converges to $x/y$; in other words: $\lim_{n \rightarrow \infty} an /b_n=(\lim_{n \rightarrow \infty} a_n)/(\lim_{n \rightarrow \infty} b_n)$

g) The sequence $(max(a_n, b_n))_{n=m}^\infty$ converges to $max (x,y)$; in other words: $\lim_{n \rightarrow \infty} max( a_n, b_n)=max(\lim_{n \rightarrow \infty} a_n,\lim_{n \rightarrow \infty} b_n)$

h)The sequence $(min(a_n, b_n))_{n=m}^\infty$ converges to $min (x,y)$; in other words: $\lim_{n \rightarrow \infty} min( a_n, b_n)=min(\lim_{n \rightarrow \infty} a_n,\lim_{n \rightarrow \infty} b_n)$

Proof:

a) We need to show that for every $\epsilon>0$ the sequence $(a_n+b_n)$ and $x+y$ are eventually $\epsilon$-close. So we fix an arbitrary $\epsilon>0$. We need to find a $N$ such that for all $n\ge N$ we have $|(a_n+b_n)-(x+y)|\le \epsilon$. However

$$|(a_n+b_n)-(x+y)| \le |a_n-x|+|b_n-y|$$

and since $(a_n)$ and $x$ are eventually $\epsilon$-close to every $\epsilon$, so they are eventually $\epsilon/2$-close; thus there is a $N_{\epsilon/2}$ such that $|a_n-x|\le \epsilon/2$ for all $n\ge N_{\epsilon/2}$. Similarly $(b_n)$ and $y$ are eventually $\epsilon/2$-close so there is a $M_{\epsilon/2}$ such that $|b_n-y|\le \epsilon/2$ for all $n\ge M_{\epsilon/2}$.

We set $N = max(N_{\epsilon/2},M_{\epsilon/2})$ and hence for each $n\ge N$, the sequences $(a_n)$ and $(b_n)$ are $\epsilon/2$-close to $x$ and $y$ respectively. Thus $|(a_n+b_n)-(x+y)| \le |a_n-x|+|b_n-y|\le \epsilon$ for every $n\ge N$.

Therefore, the sequence $(a_n+b_n)$ and $x+y$ are eventually $\epsilon$-close for each $\epsilon$ as desired.

b) We need to show that for every $\epsilon>0$ the sequence $(a_nb_n)$ and $xy$ are eventually $\epsilon$-close. We have that

$|a_n b_n-xy+b_nx-b_nx|=|b_n(a_n-x)+x (b_n-y)|\le |b_n||a_n-x|+|x| |b_n-y)|$

we need to show that $|b_n|$ is bounded.

Lemma 1: If $(c_n)_{n=m}^\infty$ is a convergent sequence, then is bounded.

Proof of lemma 1: We shall show that there is a $M$ such that $|c_n|\le M$ for each. Let $L$ be the real number such that $\lim_{n\leftarrow \infty} c_n = L$ and let $\epsilon >0$ be an arbitrary positive real number. Then there is a $N_{\epsilon}$ such that $|c_n-L|\le \epsilon$ for each $n\ge N_{\epsilon}$. Then by the triangle inequality we have $|c_n|\le |L|+\epsilon$. Now let $M= max\{\,c_m, c_m+1,...,c_{(N_{\epsilon}-1)}, |L|+\epsilon \, \}$. So if $n < N_{\epsilon}$ or $n\ge N_{\epsilon}$ the number $M \ge |c_n|$. $\Box$

By the lemma 1 it follows that $|b_n|\le M$. Now let $N= max (M,|x|)$. Then

$$|a_n b_n-xy|\le |b_n||a_n-x|+|x| |b_n-y)|\le N (|a_n-x|+|b_n-y))$$

Since the sequences $(a_n)$ and $(b_n)$ are eventually $\epsilon$-close to $x$ and $y$ for each $\epsilon$; thus they are eventually $\epsilon/2N$-close. We can choose a sufficient large $L$ such that for all $n\ge L$ $(a_n)$ and $(b_n)$ are $\epsilon/2N$-close to $x$ and $y$. Then $N (|a_n-x|+|b_n-y))\le \epsilon$ as desired.

c) We will show that for every $\epsilon>0$ the sequence $(ca_n)$ and $cx$ are eventually $\epsilon$-close. Let $\epsilon>0$ be a positive real number. Then $|c a_n -cx |\le |c| |a_n-x| $ and since $(a_n)$ converges to $x$. Then $(a_n)$ eventually $\epsilon/|c|$- close to $x$ and we're done.

d) Follows from a) and c) since $\lim(a_n-b_n)=\lim(a_n+(-1)b_n)=\lim(a_n)-\lim(b_n)$.

e) We need to show that for every $\epsilon>0$ the sequence $(b_n^{-1})$ and $x^{-1}$ are eventually $\epsilon$-close.

Lemma 2: If $(c_n)_{n=m}^\infty$ is a sequences of real number such that $L=\lim \,c_n\not=0$ and that $c_n \not=0$ for all $n\ge m$. Then $(c_n)$ is bounded away from zero, i.e., $|c_n|>\delta$ for all $n\ge m$

Proof lemma 2: Let $\epsilon>0$ and $\delta = \epsilon/2 \,$ since $(c_n)$ converges to $L$, there is a $N$ such that $d(c_n,L)\le \epsilon$ for all $n\ge N$. Then $|c_n|\le |L|+\epsilon$ by the triangle inequality. Now we set $d_n = c_n$ if $n\ge N$ and $c_n = \epsilon$ if $\,n < N$ and hence $|d_n|> \epsilon/2 = \delta$ as desired. To conclude the proof just we need to show that $(d_n)$ converges to $L$. Let us fix $\epsilon>0$

$$|d_n - L| = |(d_n-c_n)+(c_n-L)|\le |d_n-c_n|+|c_n-L| $$

it is not difficult to show that $(c_n)$ and $(d_n)$ are equivalent sequences and since $(c_n)$ is eventually $\epsilon$-close to $L$; in particular is eventually $\epsilon/2$ close to L. Thus we can pick a sufficient large $N$ such that both are at most $\epsilon/2$ and we're done. $\Box$

By lemma 2 we know that there is a $(c_n)$ which is an equivalent sequence to $(b_n)$ such that converges to $y$ and is bounded away from zero. Let c be the positive real number such that $|c_n|>c$ for each n. Then

$$ |b_n^{-1}- y^{-1}|=|c_n^{-1}- y^{-1}|= \bigg |\frac{1}{c_n}- \frac{1}{y} \bigg| = \frac {|c_n-y|}{|c_n||y|}\le \frac {|c_n-y|}{|c||y|}$$

and since $(c_n)$ is eventually $\epsilon$ close to y; thus in particular is $|c||y| \epsilon$- close to y. Thus we can pick a sufficient large $N$ such that $(c_n)$ is $|c||y| \epsilon$ close to $y$ and we're done.

f) There is nothing to prove follows from b) and e).

In the next two I have doubts about how to do it. I think I'll need a lemma to show that if one sequence is eventually bigger than the other also its limit is. But I'm not absolutely sure. So, do you think is correct my attempt of the other? and how do you think is the best way to do the last two? I would appreciate any help. Thanks in advance.

Answer 1

a) ok
b) what is the $m$ in your Lemma? Your $M$ should be taken as $M=\max\{|c_1|,\ldots,|c_{N_\epsilon}|,|L|+\epsilon\}$. Here you can choose $\epsilon=1$ to avoid new variables.
There is no need to introduce the capital $N$. Note that $|b_n||a_n-x|+|x||b_n-y|\leq 2N(|a_n-x|+|b_n-y|)$, but your idea is correct.
c) ok. Alternative: Use (b) by setting $b_n:=c$.
d) well done.
e) Is that your own proof of Lemma 2? ;) The concept of equivalent sequences is new to me, but your proofs look good.
f) yep!

g) Fix $\epsilon>0$. Then you can find an $N$ such that $x-\epsilon\leq a_n\leq x+\epsilon$ and $y-\epsilon\leq b_n\leq y+\epsilon$ for each $n\geq N$. Thus $$ \max(x,y)-\epsilon=\max(x-\epsilon,y-\epsilon)\leq\max(a_n,b_n)\leq\max(x+\epsilon,y+\epsilon)=\max(x,y)+\epsilon $$ for each $n\geq N$.

h) Follows from (g) by using $\min(s,t)=-\max(-s,-t)$ for $s,t\in\mathbb R$.

Answer 2

Hopefully there will not problem with modify the lemma 2 because is and old question: But I was thinking and it is not as useful as I thought. So, I split it in two parts.

Lemma 2.1: Every finite sequence of nonzero real numbers $c_1,c_2,...,c_n$ is bounded away from zero, i.e., $|c_i|\ge c$ for all $\,1\le i\le n$ and $c$ be a positive real number.

Proof: We prove this by induction on $n$. The base case is when $n=1$. Suppose $c_1$ is a non-zero real number, so it is clear that this sequence with one element is bounded away from zero, setting $c=|c_1|$. Now suppose that we have already proved the lemma for some natural number $n\ge 1$; we now prove it for some $n+1$, i.e., every sequence of non-zero natural numbers $c_1,c_2,...,c_n, c_{n+1}$ is bounded away from zero. Consider the sequence $c_1,c_2,...,c_n$, then by the inductive hypothesis there exists some $c>0$ such that $|c_i|\ge c$ for all $\,1\le i\le n$, i.e., it is bounded away from zero; in particular, it must be bounded away from zero by $d=\text{min}(\,c, |a_{n+1}|\,)$, notice that $d$ is a positive real number. On the other hand, $c_{n+1}$ is also bounded away from zero since $|c_{n+1}|\ge d$. Thus, there exists a $d>0$ such that $|c_i|\ge d$ for all $\,1\le i\le n+1$ and hence the sequence is bounded away from zero. This closes the induction.

Lemma 2.2: Every sequence $(c_n)_{n=m}^{\infty}$ whose elements are non-zero real numbers, and which converges to a non-zero real number $L$ is bounded away from zero, i.e., there exists a positive real number $d$ such that $|c_n|\ge d$ for all $n\ge m$.

Proof: Since $(c_n)$ is not convergent to zero, then it is not eventually close to zero for every $\epsilon>0$. So, there must be an $\varepsilon>0$ for which $(c_n)$ is not eventually $\varepsilon$-close to zero. Let us fix this $\varepsilon$. Since every convergent sequences is a Cauchy sequences so, $(c_n)$ is eventually $\varepsilon/2$ steady, i.e., there is a $N_{\varepsilon/2} \ge m$, such that $d(c_n,c_k)\le \varepsilon/2$ for all $n,k\ge N_{\varepsilon/2}$.

On the other hand, we cannot have $|c_k|\le \varepsilon$ for every $k\ge N_{\varepsilon/2}$ because this would imply that is eventually $\varepsilon$-close to zero which is a contradiction. Thus there must be a $k_0\ge N_{\varepsilon/2}$ such that $|c_{k_0}|> \varepsilon$. Now since we already know that $d(c_n,c_{k_0})\le \varepsilon/2$, for every $n\ge N_{\varepsilon/2}$, using the triangle inequality we thus conclude $|c_n|\ge \varepsilon/2$ for each $n\ge N_{\varepsilon/2}$.

Then, $(c_n)$ is bounded away from zero when $n\ge N_{\varepsilon/2}$. But the case when $n< N_{\varepsilon/2}$ follows immediately from lemma 2.1, because is a sequences of non-zero elements and it is finite. So, we thus have $|c_n|\ge c>0$ when $n< N_{\varepsilon/2}$. To conclude the proof, we set $\,d=\text{min}(\,\varepsilon/2,c\,)$. And since $d>0$ because both $\,c,\varepsilon/2>0$ and $|c_n|\ge d$ for all $n\ge m$ (splitting into two cases: when $n< N_{\varepsilon/2}$ and $n\ge N_{\varepsilon/2}$) we thus have $(c_n)$ is bounded away from zero as desired.