Why does $f(x)=x$ not have a maximum on open interval $(0,1)$?

Question

On a closed interval $[0,1]$ it evidently has $\max$ at $1$ - so wouldn't the $\max$ for the open interval just be whatever decimal number is just before $1$?

Answer 1

Assume you would have such a decimal number, lets call it $x$. Then clearly $x<1$.

Now take a look at $\frac{1+x}{2}$. You observe $\frac{1+x}{2}<1$ and $\frac{1+x}{2}>x$, thus this number is still in $(0,1)$ but larger than $x$ and this yields a contradiction, since $x$ could not have been such a decimal number.

Answer 2

wouldn't the max for the open interval just be whatever decimal number is just before 1?

Yes, if there existed a number "just before $1$", then that number would be the maximum of $f$.

But such a number doesn't exist.

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To actually prove that $f$ does not have a maximum, you can prove the statement

For every $x\in(0,1)$, $f(x)<f\left(\frac{x+1}{2}\right)$ and $\frac{x+1}{2}\in (0,1)$.

from which the fact that

for every $x\in(0,1)$, $x$ is not a maximum of $f$

clearly follows.

Answer 3

You use there words HAS a max and THE max yourself.

The concept that works for "just before one and such" is a Supremum which is much much neater.

Every max is a sup, but a sup need not be a max: a sup need not be a value attained in the domain.

But a sup has the property that it is closer to $1$ in this case than any other upper bound. Every upper bound is either greater than the supremum or equal to it.

It is the value in the LIMIT of $x$ getting closer and closer to $1$ regardless of if it is defined in $1$ or defined just up to $1$ as in the case of the open set $(0,1)$.

But calculus:

The derivative of the function $f$ that is defined in the open interval is $$\frac{df}{dx}:(0,1) \to \mathbb{R}$$ $$\frac{df}{dx}(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}= \lim_{h\to 0} \frac{x+h-x}{h}= \lim_{h\to 0} \frac{h}{h} = \lim_{h\to 0} 1=1$$

Which makes sense anyways, except when at $0$ and $1$. Luckily excluded.

This means that as $x$ changes, in whichever way, $f$ smoothly changes in exactly the same way, because it is a mapping $x \mapsto x$.

Another important property is of course that $f$ is continuous. This has to do with open sets, and with considering $f$ as mapping sets and not so much points and numbers: $$f^{-1}:(0,1) \to (0,1)$$ the inverse of a function is for one technically nicer. if open sets virtually portrayed by the action of $f$ into the target set are mapped back to the original domain then the result is always an open subset of the domain. $f$ is also continuous and differentiable, and there's a theorem about local extremum points in open sets that says that the derivative must be exactly $0$ at these points (maximum points, minimum points) of $f$. But we got $1$ for the derivative everywhere in the open set. so there is no maximum..