# Trouble putting the finishing touches on the proof of the squeeze theorem.

I want to prove the following claim:

$$\Big (a \leq b \leq c \land \lvert a - L \rvert \lt \epsilon \land \lvert c - L \rvert \lt \epsilon \Big) \rightarrow \lvert b - L \rvert \lt \epsilon$$

I think I am on the correct track, so, if possible, I would appreciate if responders could "finish off" where I am stuck (or provide hints). If I am not on the correct path, feel free to abandon my approach.

Relevant lemma:

$$a \lt b \lt c \rightarrow \big ( \lvert b \rvert \lt \lvert c\rvert \ \lor \lvert b \rvert\lt \lvert a \rvert \big)$$

There are two cases: $$b \geq 0$$ or $$b \lt 0$$.

Case 1: $$b \geq 0$$

By assumption, $$b\lt c$$. Therefore, $$0 \leq b \lt c$$. By definition of absolute value, $$\lvert b \rvert=b$$ and $$\lvert c \rvert =c$$. Thus, $$\lvert b \rvert \lt \lvert c \rvert$$.

Case 2: $$b \lt 0$$

By assumption, $$a \lt b$$. Therefore, $$a \lt b \leq 0$$. By definition of absolute value, $$\lvert b \rvert = -b$$ and $$\lvert a \rvert =-a$$. Multiplying our previous inequality by $$-1$$, we have: $$-a \gt -b \geq 0$$. Thus, $$\lvert a \rvert \gt \lvert b \rvert$$ (or $$\lvert b \rvert \lt \lvert a \rvert$$). $$\square$$

Now, if $$\lvert a - L \rvert \lt \epsilon$$, then we have $$\lvert a \rvert - \lvert L \rvert \leq \lvert a - L \rvert \lt \epsilon$$ (reverse triangle inequality). This implies that $$\lvert a \rvert \lt \epsilon + \lvert L \rvert$$.

A similar argument will show that $$\lvert c \rvert \lt \epsilon + \lvert L \rvert$$.

Combining this with our lemma, we have two cases:

Case 1: $$\lvert b \rvert \lt \lvert c \rvert$$

In this case, $$\lvert b \rvert \lt \lvert c \rvert \lt \epsilon + \lvert L \rvert$$. This implies that $$\lvert b \rvert - \lvert L \rvert \lt \epsilon$$. Unsure of how to finish

Case 2: $$\lvert b \rvert \lt \lvert a \rvert$$

In this case, $$\lvert b \rvert \lt \lvert a \rvert \lt \epsilon + \lvert L \rvert$$. This implies that $$\lvert b \rvert - \lvert L \rvert \lt \epsilon$$. Unsure of how to finish

I originally thought that proving $$\lvert b \rvert - \lvert L \rvert \lt \epsilon$$ would let me usefully invoke the reverse triangle inequality to conclude that $$\lvert b - L \rvert \lt \epsilon$$...but I now realize that $$\lvert b \rvert - \lvert L \rvert \lt \epsilon$$ and $$\lvert b \rvert - \lvert L \rvert \leq \lvert b - L \rvert$$ does not imply that $$\lvert b - L \rvert \lt \epsilon$$.

Any suggestions are greatly appreciated.

• Your approach looks a bit complicated. I would argue that $b-L \le c - L < \epsilon$ and $b - L \ge a - L > -\epsilon$. Apr 15 '21 at 4:59

$$a\le b \le c$$ and $$|a-L|<\epsilon$$ and $$|c-L|<\epsilon$$
$$\implies -\epsilon< a-L\le b-L\le c-L<\epsilon$$
So since epsilon is positive $$|b-L|<\epsilon$$
If you can suppose that $$a\leq b \leq c$$ and that $$\lvert a-L\rvert<\epsilon$$ and $$\lvert c-L\rvert<\epsilon$$. Then "opening up" the absolute value inequality results in $$-\epsilon < a- L < \epsilon$$ and $$-\epsilon < c- L < \epsilon$$. Then since $$a\leq b\leq c$$ it's also true that $$a-L \leq b-L \leq c-L$$ thus by transitivity $$-\epsilon < a- L < b-L < c-L < \epsilon$$.
Here the main take away is $$-\epsilon < b- L < \epsilon$$ thus by closing the absolute value again, $$\lvert b-L\rvert<\epsilon$$.