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Let $f$ be a real-valued function defined on $[0,1]$ such that $f(0)>0, f(x) \ne x ~\forall x,$ and $f(x) \leq f(y)$ whenever $x \le y.$ Let $A =\{x : f(x)>x\}.$ Prove that $\sup A \in A.$

Attempt: Let us suppose that $\sup A = a \notin A.$ Then $f(a) \leq a$. But, since, $f(x) \ne x \implies f(a) < a~~~~ ..........(1)$

Now, by the definition of supremum, every open ball $B(a,r) $ contains at least one element $y$ of $A$ . Then, $y \in B(a,r)$ .

Since, $y \in A \implies f(y) > y ~~......(2)$

$ y < a$ and $f$ is increasing $ \implies f(y) \leq f(a)~~~~~~........(3)$

From $(1),(2),(3) : y <f(y) \leq f(a)<a$

How do I bring some contradiction now?

Thank you for your help..

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  • $\begingroup$ There is no function $f$ satisfying $f(x) < f(y)$ whenever $x \le y.$ $\endgroup$
    – Umberto P.
    Oct 12, 2014 at 12:35
  • $\begingroup$ I have corrected it :-) $\endgroup$
    – MathMan
    Oct 12, 2014 at 12:37
  • $\begingroup$ @UmbertoP. The empty function satisfies it :P $\endgroup$
    – Nishant
    Oct 12, 2014 at 13:14

3 Answers 3

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Direct proof: for any $x\in A$, we have $$ f(\sup A)\geq f(x)>x. $$ The first inequality uses monotonicity and the second uses $x\in A$. This means that $f(\sup A)$ is an upperbound for $A$. In particular, $f(\sup A)\geq\sup A$. But $f(\sup A)\neq \sup A$, so we must have $f(\sup A)>\sup A$. That is, $\sup A\in A$.

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    $\begingroup$ Wow. Thank you very much for the answer. $\endgroup$
    – MathMan
    Oct 12, 2014 at 13:31
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From your work, you get $y < f(a) < a$ for all $y \in A$ sufficiently close to $a$ (i.e. in some ball $B(a,r)$). Taking supremums over $A$, we see that $$ a = \sup A = \sup_{y \in A} y \le \sup_{y \in A} f(a) \le \sup_{y \in A} a = a. $$ Therefore $a = f(a)$, a contradiction by (1).

Hope that helps,

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  • $\begingroup$ Thank you very much for the answer. $\endgroup$
    – MathMan
    Oct 12, 2014 at 13:31
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You want to use that there is such a $y$ for each $r > 0$, and that then $a - r < y_r$. Together with what you got you end up with:

$\forall r > 0 \ \left ( a - r < f(a) < a \right )$

and this yields the contradiction.

Edit: To see the latter claim, pick $r := a - f(a)$. As $f(a) < a$, this satifies $r > 0$. Then $a - r < f(a)$ evaluates to $f(a) < f(a)$, which is a contradiction.

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    $\begingroup$ Thank you for the answer. If $y = a-r$, we we have $y < f(a) < a$. But how does this violate the order structure of $\mathbb R$? $\endgroup$
    – MathMan
    Oct 12, 2014 at 12:47
  • $\begingroup$ I've added a clarification. Note that you cannot assume that $y = a - r$ for two reasons: One, $y$ is inside the ball $B(a,r)$, not on the boundary. Two, you only that there are points inside the intersection of the ball and $A$, but you cannot assume that there are no points inside the ball and not in $A$. $\endgroup$
    – Arno
    Oct 12, 2014 at 13:06
  • $\begingroup$ Ohkay .. Thank you for your answer :-) $\endgroup$
    – MathMan
    Oct 12, 2014 at 13:33

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