# Thread: Is this proof correct?

1. ## Is this proof correct?

prove that if f(x) is monotone increasing on an open interval (a,b) and x_0 is any point in (a,b), then lim_{x \to x_0+} f(x) exists and satisfies lim_{x to x_0+} f(x) >= f(x_0). (from the right side)

Is this proof correct? if not could someone please correct it/ explain what I'm doing wrong?

Fix $x_0$ in (a,b). First consider x in ( $x_0$,b] then f(x)>= f( $x_0$) so inf f(x) ( $x_0$<x<b) >=f( $x_0$)

Claim: lim f(x)=inf f(x) ( $x_0$<=x<b) . Let S=inf f(x). Let $\epsilon$>0. Then there exists x $\epsilon$ in ( $x_0$,b) such that f(x $\epsilon$)<S- $\epsilon$.

But if x is in (x $\epsilon$, $x_0$) then f(x $\epsilon$)>=f(x) so S<=f(x)<S- $\epsilon$

Hence |f(x)-s|< $\epsilon$ for all x in (x $\epsilon$, $x_0$).

therefore, lim f(x) (x-> $x_0$+) exists and equals S since S>=f( $x_0$) we get lim (x-> $x_0$+) >= f( $x_0$)

2. Frankly, I find your post very hard to read. Let me give you some alternative ways to see this.

If $a then let $f(c+)=\inf\{f(x):x\in (c,b]\}$ and $f(c-)=\{\sup\{f(x):x\in [a,c)\}$.
It is not hard the see that for an increasing function each of these exists.

Clearly $\lim _{x \to c^ + } f(x) = f(c + ) \ge f(c)$ and likewise Clearly $\lim _{x \to c^ - } f(x) = f(c - ) \le f(c)$.

I hope this helps you with notation.
I will be glad to help with questions you may have.

3. I think the property yo uare using in your wrong has a little mistake:

Claim: lim f(x)=inf f(x) (<=x<b) . Let S=inf f(x). Let >0. Then there exists x in (,b) such that f(x)<S-.
I think it should say:

For ever \epsilon there exists $x_\epsilon \in (x_0,b)$ such that $f(x_\epsilon) <\inf\{f(x):x_0\leq x \leq b \}+\epsilon$

what do you think?

4. Originally Posted by tn11631
Whatever you said..
I think you mean to prove that if $fa,b)\mapsto\mathbb{R}" alt="fa,b)\mapsto\mathbb{R}" /> is increasing then $\lim_{x\to x_0^-}f(x)=\sup_{x\in (a,x_0)}f(x)$. To see this let $\varepsilon>0$ be given. By virtue of the definition of supremum we know that there exists some $f(\xi)\in f((a,x_0))$ such that $\sup_{x\in(a,x_0)}f(x)-\varepsilon. Notice though by definition we have that $\sup_{x\in(a,x_0)}f(x)-f(\xi)=\left|\sup_{x\in(a,x_0)}f(x)-f(\xi)\right|$. But, by $f$'s monotonicity we have that for any $\xi<\xi'<\xi$ that $f(\xi). In particular if $x\in(\xi,x_0)\implies \left|\sup_{x\in(a,x_0)}f(x)-f(x)\right|=\sup_{x\in(a,x_0)}f(x)-f(x)<\sup_{x\in(a,x_0)}f(x)-f(\xi)<\varepsilon$. It follows that $\lim_{x\to x_0^-}f(x)=\sup_{x\in(a,x_0)}f(x)$.