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Given: $(a_n), (b_n)$, and $(c_n)$ are sequences, with $a_n \le b_n \le c_n$ for all n. Also, $a_n \to a$ and $c_n \to a$.

Prove by contradiction: $(b_n)$ converges and $b_n \to a$.

Here is my attempt. Please let me know if this is a viable proof, and how I can improve upon it.

Proof:

Suppose $(b_n)$ does not converge to $a$.

Then, $\exists \epsilon _b \gt 0$ $ \forall M \in \mathbb N$ such that $\forall n \ge M, |b_n-a| \ge \epsilon _b$

and $\forall \epsilon \gt 0, \exists N_1, N_2 \in \mathbb N$ such that $\forall n \ge N_1, |c_n-a| \lt \epsilon$ and $\forall n \ge N_2, |a_n-a| \lt \epsilon$.

Let $N=max(N_1,N_2)$.

Letting $M=N$, we get:

$\exists \epsilon _b \gt 0$ such that $\forall n \ge N$, $|b_n-a| \ge \epsilon _b, |c_n-a| \lt \epsilon _b, |a_n-a| \lt \epsilon _b$.

Thus:

$a-\epsilon _b \lt a_n $

$a-\epsilon _b \gt c_n$

$a-\epsilon _b \ge b_n$ or $a+\epsilon _b \le b_n$

As a result, $\exists \epsilon _b \gt 0$ such that $\forall n \ge N, b_n \le a-\epsilon _b \lt a_n$ or $c_n \lt a + \epsilon _b \le b_n$, contradicting the fact that $a_n \le b_n \le c_n$. Therefore, $b_n$ must converge to $a$.

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  • $\begingroup$ This should help math.stackexchange.com/questions/1135350/… $\endgroup$ – Breton Thomas Sep 22 '16 at 0:43
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    $\begingroup$ $\exists \epsilon _b \gt 0$ $ \forall M \in \mathbb N$ such that $\forall n \ge M, |b_n-a| \ge \epsilon _b$ is wrong. You are trying to negate $\forall \epsilon _b \gt 0$ $ \exists M \in \mathbb N$ such that $\forall n \ge M, |b_n-a| < \epsilon _b$. You should get $\exists \epsilon _b \gt 0$ $ \forall M \in \mathbb N$ such that $\exists n \ge M, |b_n-a| \ge \epsilon _b$. (Specifically your $\forall n$ should be $\exists n$.) $\endgroup$ – James Sep 22 '16 at 0:58
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I think that your initiating implication is invalid:

Suppose $b_n$ does not converge to $a$.

Then, $\exists \epsilon _b \gt 0$ $ \forall M \in \mathbb N$ such that $\forall n \ge M, |b_n-a| \ge \epsilon _b$

A non-converging sequence $b_n$ might well have infinite points arbitrarily close to $a$ (without converging to it).

  • A (direct) proof to the Squeeze theorem can go like this:

Proof: Since $a_n \leq b_n \leq c_n$ then $0\leq b_n-a_n\leq c_n-a_n$, thus $|b_n-a_n|\leq c_n-a_n$.

Combining the above with the fact that $\lim(c_n-a_n)=a-a=0$ we get: $\lim(b_n-a_n)=0$.

Now we can write the terms of $(b_n)$ as the sum of the terms of two converging sequences: $b_n=(b_n-a_n)+a_n$, so we have: $$ \lim b_n=\lim\big((b_n-a_n)+a_n \big)=\lim(b_n-a_n)+\lim a_n=0+a=a $$

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If the inequality is true, you get that if $b_n<a$ at that point, then $b_n<a_n$ a contradiction. Furthermore, if $b_n>a$ then $b_n>c_n$ a further contradiction. So, if it weren't accurate that $b_n\to a$ at that point, it falsify's the statement about the inequality ( I originally said invalidates, but in philosophy from khan academy, they go over that statements are either true or false, arguments are valid or invalid).

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KonKans answer already points to the mistake in your argument.

The following proof by contradiction can easily be converted into an even simpler direct proof:

If the $b_n$ do not converge to $a$ there is an $\epsilon_0>0$ such that there are arbitrarily large $n$ with $|b_n-a|\geq\epsilon_0$. On the other hand, there is an $n_0$ such that $|a_n-a|<\epsilon_0$ and $|c_n-a|<\epsilon_0$ for all $n>n_0$. It follows that for all large $n$ we have $$a-\epsilon_0<a_n\leq b_n\leq c_n<a+\epsilon_0\ ,$$ which excludes the occurrence of "bad" $b_n$s.

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Consider the set $A_n=[a_n,c_n]$, $n=1,2, \dots$

$$\limsup A_n=\bigcap_{n-1}^{\infty} \bigcup_{n\geq m} A_m=\{a\}\space\text{(Why?)}$$

$$\liminf A_n=\bigcup_{n-1}^{\infty} \bigcap_{n\geq m} A_m=\{a\}\space\text{(Why?)}$$

and $b_n\in A_n$, for all $n\in \mathbb{N}$.

Hence $\lim_{n\to\infty} b_n=a$

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