1. ## Monotone Functions

I have attempted this proof and would appreciate any feedback. Thanks.

Let $\displaystyle I$ be some interval. If $\displaystyle f:I\rightarrow \mathbb{R}$ is increasing and $\displaystyle a\in I$, prove that $\displaystyle f(a)$ is an upper bound for $\displaystyle \{f(x)|x<a\}$ and $\displaystyle lim_{x\nearrow a}f(x)=sup\{f(x)|x<a\}$.

Proof: Let $\displaystyle f$ be an increasing function. Then, by definition, for all $\displaystyle x,y\in I, x<y\Rightarrow f(x)<f(y)$. Then, let $\displaystyle x<a$ for $\displaystyle a\in I$ and all $\displaystyle x\in I$. Thus, since $\displaystyle x<a\Rightarrow f(x)<f(a)$ for all $\displaystyle a,x\in I$. Hence, $\displaystyle a$ is an upper bound for $\displaystyle \{f(x)|x<a\}$.

Then, since $\displaystyle x<a$ for all $\displaystyle x,a\in I$, then $\displaystyle a$ must be the least upper bound for $\displaystyle \{f(x)|x<a\}$. Hence, $\displaystyle lim_{x\nearrow a}f(x)=sup\{f(x)|x<a\}$.

2. You have some confused notation here.
You have $\displaystyle a$ in two different ways.
What you want is to note that $\displaystyle f(a)$ is an upper bound for $\displaystyle \{f(x):x<a\}$.
Now consider $\displaystyle A=\sup\{f(x):x<a\}$ so that $\displaystyle A\le f(a)$.
Then show that $\displaystyle \displaystyle \lim _{x \to a^ - } f(x) = A$

3. Major remarks.

Hence, $\displaystyle a$ is an upper bound for $\displaystyle \{f(x)|x<a\}$.
Should be f(a), not a.

The second claim is not really proved.
Then, since $\displaystyle x<a$ for all $\displaystyle x,a\in I$, then $\displaystyle a$ must be the least upper bound for $\displaystyle \{f(x)|x<a\}$.
Even the claim that f(a) (rather than a) is the least upper bound for $\displaystyle \{f(x)\mid x<a\}$ is incorrect. Consider

$\displaystyle f(x)= \begin{cases} x & x<a\\ x+1 & x\ge a \end{cases}$

Then $\displaystyle \sup\{f(x)\mid x<a\}=a$, but $\displaystyle f(a)=a+1$.

Even if $\displaystyle f(a)=\sup\{f(x)\mid x<a\}$, how does it help prove the following?
Hence, $\displaystyle lim_{x\nearrow a}f(x)=sup\{f(x)|x<a\}$.
After all, $\displaystyle \lim_{x\nearrow a}f(x)=f(a)$ only if f is left-continuous at a. In my opinion, there is at least not enough details here that are expected from a proof of such a simple fact.

Minor remarks.

Then, let $\displaystyle x<a$ for $\displaystyle a\in I$ and all $\displaystyle x\in I$.
You can't expect that x < a for all x in I. Since a is already fixed in the problem statement, this phrase should say, "Let x < a for some x in I".

Thus, since $\displaystyle x<a\Rightarrow f(x)<f(a)$ for all $\displaystyle a,x\in I$.
Again, it is not the case that f(x) < f(a) holds for all a, x in I.

4. Alright, so from what I gather, most of the first part was ok besides the fact that $\displaystyle f(a)$ is the upper bound and not $\displaystyle a$.

As for the second part...
Let $\displaystyle A=\sup\{f(x):x<a\}$ and $\displaystyle A\leq f(a)$. Then,
since $\displaystyle f(a)$ is an upper bound for $\displaystyle \{f(x)|x<a\}$ and f is increasing then there is no $\displaystyle x>a$ such that $\displaystyle f(x)>f(a)$. Thus, $\displaystyle lim_{x\rightarrow a^-}f(x)\leq f(a)$. If we let this limit be called $\displaystyle A$ then we have that $\displaystyle lim_{x\rightarrow a^-}f(x)=A$ where $\displaystyle A\leq f(a)$. Hence, we have that $\displaystyle lim_{x\nearrow a}f(x)=sup\{f(x)|x<a\}$.

I'm not sure this is very strong but its my second attempt.

5. Sorry, too many flaws. The second part of the problem -- proving $\displaystyle \lim_{x\to a^-}f(x)=\sup\{f(x)\mid x<a\}$ -- has nothing to do with f(a). In the example I gave above, f(a) is different from $\displaystyle \sup\{f(x)\mid x<a\}$.

Let $\displaystyle A=\sup\{f(x):x<a\}$ and $\displaystyle A\leq f(a)$.
Are you assuming $\displaystyle A\leq f(a)$? In this case, since this assumption is not in the problem statement, you later need to consider the possibility that $\displaystyle A>f(a)$. However, I thought that you have proved this in part 1. Therefore, you should say, "Let $\displaystyle A=\sup\{f(x):x<a\}$. We have already shown that $\displaystyle A\leq f(a)$." (To repeat, the second sentence is irrelevant.)

Then, since $\displaystyle f(a)$ is an upper bound for $\displaystyle \{f(x)|x<a\}$ and f is increasing then there is no $\displaystyle x>a$ such that $\displaystyle f(x)>f(a)$.
What happens to the right of $\displaystyle a$ is irrelevant to the problem.

Thus, $\displaystyle lim_{x\rightarrow a^-}f(x)\leq f(a)$.
I don't see how this follows from the previous discussion.

If we let this limit be called $\displaystyle A$
This is redefining A.

then we have that $\displaystyle \lim_{x\rightarrow a^-}f(x)=A$ where $\displaystyle A\leq f(a)$. Hence, we have that $\displaystyle \lim_{x\nearrow a}f(x)=sup\{f(x)|x<a\}$.
It looks like this conclusion is based on confusing two meanings of A.

Let us define $\displaystyle A=\sup\{f(x)\mid x<a\}$, as you have done. Here is how I would prove that $\displaystyle \lim_{x\to a^-}f(x)=A$. We show that for every $\displaystyle \varepsilon>0$ there exists a $\displaystyle \delta>0$ such that when $\displaystyle a-\delta<x<a$ and $\displaystyle x\in I$ we have $\displaystyle A-\varepsilon<f(x)\le A$. Note that the second inequality, i.e., $\displaystyle f(x)\le A$, holds by the definition of $\displaystyle \sup$.

Let $\displaystyle \varepsilon$ be given. Also by the definition of $\displaystyle \sup$, $\displaystyle A-\varepsilon$ is not an upper bound of $\displaystyle \{f(x)\mid x<a\}$; therefore, there exists an $\displaystyle x_0<a$ such that $\displaystyle f(x_0)>A-\varepsilon$. But then, since f is increasing, for every $\displaystyle x$ such that $\displaystyle x_0<x<a$ we have $\displaystyle A-\varepsilon<f(x_0)<f(x)$. Therefore, we can take $\displaystyle \delta=a-x_0$.

6. A quasi-continuous function is one for which the right and left hand limits exist at each point. It is a standard theorem in analysis that any monotonic function is quasi-continuous. That is the essence of this problem. You are asked to show that the limit on the left exist at each point.