# Prove the Squeeze Theorem (for limits of sequences).

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 $a$.

• This should help math.stackexchange.com/questions/1135350/… – Breton Thomas Sep 22 '16 at 0:43
• $\exists \epsilon _b \gt 0$ $\forall M \in \mathbb N$ 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 $\exists n$.) – James Sep 22 '16 at 0:58

## 4 Answers

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: $$\lim b_n=\lim\big((b_n-a_n)+a_n \big)=\lim(b_n-a_n)+\lim a_n=0+a=a$$

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).

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.

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