On the use of the "min" function / operator in epsilon-delta limit proofs: how does " delta = min {a,b}" guarantee " delta <a AND delta < b"?

Question

Below a limit proof that requires in the end the use of the " min" function.

I know I have to use " delta = min {a,b} " but I do not understand clearly how the " min function" does the job.

Can it be shown formally that " delta = min { 1, epsilon/15} " guarantees that : "( delta < 1 AND delta < epsilon /15)".

My question : how to explain the step from (8) to (9) in the proof below?

Also: when checking a delta value in the form of "min { a, b}", do I have to proceed to a double verification, one for a, and one for b?

Answer 1

To be precise, $\delta = \min(a, b)$ means $\delta \le a$ and $\delta \le b$.

As a programming construct, min can be defined as min(a, b) = if a $\le$ b then a else b.

So any proofs involving min do have to consider all possibilities.

The inequality in the definition can be either $<$ or $\le$ - it does not matter since the result is the same if $a = b$.

The result is that $c = \min(a, b) \implies (c \le a) \land (c \le b) $.

Answer 2

Either $a \leq b$ or $b \leq a$. The minimum is equal to one and less than the other. So it's less than or equal to both.

If $|x-1| < \delta$, then either: $$ |x - 1 | < \delta = a < b $$ or $$ |x - 1 | < \delta = b < a $$ Either way, $|x-1| \leq a$ and $|x-1| \leq b$ both hold.