Help with a theorem about continuous increasing functions I'm having a lot of trouble understanding the following theorem from my analysis class.
Let $I$ be a subset of the real numbers and let $f : I \to R$ be increasing on $I$.
Suppose that $c$ is an element of $I$ but not an endpoint of $I$.
Then:


*

*$\lim_{x \to c^-}f(x) = \sup\, \{f(x): x \in I, x < c\}$

*$\lim_{x \to c^+}f(x) = \inf\, \{f(x): x \in I, x > c\}$


I really don't understand what this theorem is stating.  This theorem is in a chapter about continuous functions, section titled "Monotone and increasing functions".  It follows a review about what monotone functions are and definitions of increasing and decreasing functions.  Any help is appreciated.
 A: The continuity of $f$ is not given. We are however given that $f$ is increasing in $I$. Let $c \in I$ and $c$ not be an end point of $I$. This condition of $c$ not being an end point, but an interior point of $I$ is given so that we have points $x \in I$ which are less than $c$ and points $x \in I$ which are greater than $c$.
I will provide the proof for $\lim_{x \to c^{-}}f(x) = \sup\{f(x)\mid x \in I, x < c\}$. The other result can be handled similarly. First of all note that if $x < c, x \in I$ then $f(x) < f(c)$ so that the set $A = \{f(x)\mid x \in I, x < c\}$ is bounded above by $f(c)$ and hence $\sup A$ exists by the completeness property of real number system.
To show that $\lim_{x \to c^{-}}f(x) = \sup A$ we need to show that for any $\epsilon > 0$ there exists a $\delta > 0$ such that $0 < c - x < \delta$ implies $|f(x) - \sup A| < \epsilon$. This is easy to establish because $f$ is increasing on $I$. By the definition of supremum we have at least one element say $f(x') \in A$ such that $f(x') > \sup A - \epsilon$. By definition of set $A$, we have $x' < c$ and let $\delta = c - x' > 0$. Now consider any point $x$ which satisfies $ 0 < c - x < \delta = c - x'$. Then we have $x' < x < c$ and therefore $\sup A - \epsilon < f(x') \leq f(x) \leq \sup A$. This means that $|f(x) - \sup A| = \sup A - f(x) < \epsilon$.
Thus the definition of the limit is satisfied and we have established that $\lim_{x \to c^{-}}f(x) = \sup A$.
Informally as $x \to c^{-}$, $x$ is approaching $c$ by taking values less than $c$ and increasing bit by bit reaching upto $c$. In turn the function $f(x)$ is also increasing steadily but always remaining less than or equal to $f(c)$ (i.e. taking values in set $A$) and finally reaches $\sup A$ in the limit.
The question which you have asked is a very important property of monotone functions. Monotone functions possess left and right hand limits at each point precisely because of their monotonicity.
Update: I hope you have studied the concepts of supremum and infimum before reaching to this theorem. If not then it is better that you first get acquainted with these concepts. I provide a definition in plain English without too much symbolism.
A number $M$ is said to be the supremum of a non-empty set $A \subseteq \mathbb{R}$
if 1) no member of $A$ exceeds $M$ and 2) every number less than $M$ is exceeded by at least one member of $A$.
A number $m$ is said to be the infimum of a non-empty set $A \subseteq \mathbb{R}$ if 1) $m$ does not exceed any member of $A$ and 2) every number greater than $m$ exceeds at least one member of $A$.
The completeness property of real number system states that a non empty set of numbers which is bounded above possesses a supremum and using this one can show that a non-empty set of numbers which is bounded below possesses an infimum.
A: Since $I$ is a closed interval (namely $[a,b]$) and $f$ is increasing, then the image of $f$ has an upper bound and so the sup does exists. Let $S$ be the sup. To show that $S=\lim_{x\to c_-} f(x)$ you have to show that for any $\epsilon>0$ there exists $\delta>0$ such that $$0<|c-x|=c-x<\delta\Rightarrow 0<|f(x)-S|<\epsilon.$$
A: a subset of the reals is too broad a class to be interesting. one wants the subset at least to be dense in some interval. clearly monotonicity is quite a strong condition, so the question which naturally presents itself is how far a monotone function can be from being continuous. the answer is that a monotone function cannot be everywhere discontinuous on an interval of the reals, but (and this is what is intriguing) it can be discontinuous on a countable dense subset. hope this is of some use to you.
A: When $I$ is an arbitrary subset of the reals the assumption "$\>c$ is not an endpoint of $I\>$"  doesn't make much sense. What is essential is that any (half) neighborhood of $c$ intersects $I$. Therefore assume that $f:\>I\to{\mathbb R}$ is increasing and that an arbitrary $c$, $\>-\infty\leq c\leq\infty$, is given. Then we can say the following:
If any right-hand neighborhood of $c$ intersects $I$ then
$$\lim_{x\to c+}f(x)=\inf\,\{f(x)|x\in I, \  x>c\}\geq-\infty\ ,$$
and if any left-hand neighborhood of $c$ intersects $I$ then
$$\lim_{x\to c-}f(x)=\sup\,\{f(x)|x\in I, \  x<c\}\leq\infty\ .$$
It suffices to prove the first statement. Assume that $$\inf\,\{f(x)| x>c\}=\alpha\in{\mathbb R}\ ,$$ and let an $\epsilon>0$ be given. We have to prove that there is a right-hand neighborhood $U:=\ ]c, d[\ $ of $c$ such that
$$|f(x)-\alpha|<\epsilon\qquad\forall x\in U\cap I\ .\tag{1}$$
By assumption on $I$ and definition of  $\alpha$ there is a point $d\in I$ with $d>c$ and $f(d)<\alpha+\epsilon$. Since $f$ is increasing we then have $f(x)<\alpha+\epsilon$ for all $x\in U$, and by definition of $\alpha$ we certainly have $f(x)\geq\alpha$ for all $x\in U\cap I$. This proves that $(1)$ holds.
The case $\alpha=-\infty$ is handled similarly; it can not occur when there are  points of $I$ to the left of $c$.
