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.

  • 2
    $\begingroup$ Your approach looks a bit complicated. I would argue that $b-L \le c - L < \epsilon$ and $b - L \ge a - L > -\epsilon$. $\endgroup$
    – Martin R
    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, $\lvert b-L\rvert<\epsilon$.


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