Prove f has a fixed point if it is increasing but not necessarily continuous Suppose $f : [0, 1] \rightarrow [0, 1]$ is increasing (but not necessarily continuous).
Show that there is a number $x \in [0, 1]$ with $f(x) = x$.
(Hint: You can’t apply the IVT directly because the function need not be
continuous. Draw a picture and try to copy the proof of the Intermediate
Value Theorem.) 
I don't understand how copying the IVT and connecting it to my graph would help me prove this since the IVT has nothing to do with increasing functions(or am I wrong about this?)
 A: Hint
What about $$c=\inf\{x\in [0,1]\mid f(x)\leq x\}$$ if it exist ?
A: Similar to Mark's solution. It makes use of Completeness principle.
If $f(0) = 0$, then $x = 0$; if $f(1) = 1$, then $x = 1$; otherwise let $a_0=f(0)$, $b_0=f(1)$, and, for $n=1,2,\dots$ repeat the following.
(1) Let $\Delta_n =\frac{a_{n-1}+b_{n-1}}{2}$,
(2) If $f\left(\Delta_n\right) = \Delta_n$, then $x = \Delta_n$ and we are done.
(3) If $f\left(\Delta_n\right) < \Delta_n$, set $a_n = a_{n-1}$, $b_n = \Delta_n$.
(4) If $f\left(\Delta_n\right) > \Delta_n$, set $a_n = \Delta_n$, $b_n = b_{n-1}$.
This leads to the construction of a monotonically increasing sequence $\left(a_n\right)_{n\in \mathbb Z^+}$ and a monotonically decreasing sequence $\left(b_n\right)_{n\in \mathbb Z^+}$, such that
$$0=a_0 \leq a_1 \leq \cdots \leq a_n \cdots \leq b_n \cdots \leq b_1 \leq b_0=1,$$
with
\begin{equation}
f(b_n)<b_n,\tag{1}\label{eq:down}
\end{equation}
and
\begin{equation}
f(a_n)>a_n.\tag{2}\label{eq:up}
\end{equation}
Furthermore we have
\begin{equation}b_n - a_n \leq \frac{1}{2^n}.\tag{3}\label{eq:diff}\end{equation}
So, by boundedness and monotonicity the two sequences converge (here Completeness comes into play), and by \eqref{eq:diff}, they converge to the same number $\alpha \in [0,1]$. 
Suppose now $f(\alpha) > \alpha$. Then, for sufficiently large $n$,
$$\alpha \leq b_n < f(\alpha),$$
which, together with \eqref{eq:down} gives
$$f(b_n) < b_n < f(\alpha),$$
contradicting monotonicity of $f$. Similarly, if $f(\alpha) < \alpha$, then, for sufficiently large $n$,
$$f(\alpha) < a_n \leq \alpha,$$
so that, using \eqref{eq:up},
the inequality
$$ f(a_n) > a_n > f(\alpha)$$
leads again to a contradiction. Thus, it must be $f(\alpha) = \alpha$. 

Note
Completeness is necessary for the statement to hold true. As a counterexample consider the function $f: [0,1]\cap \Bbb Q \rightarrow [0,1]\cap \Bbb Q$
$$f(x) = -\frac{1}{x-3}.$$
This function is continuous and strictly increasing in its domain, but there is no point in the domain in which $f(x) = x$.
A: Start by defining the sequences $l_n, u_n$, where the variable names stand for "lower" and "upper". And let $x_n \equiv (l_n + u_n)/2$.
$l_0 \equiv 0, u_0 \equiv 1$
Now inductively define $l_k, u_k$:
If $f(x_{k-1}) \lt x_{k-1}$, define $u_k \equiv x_{k-1}, l_k \equiv l_{k-1}$ (shift the upper bound down).
If $f(x_{k-1}) > x_{k-1}$, define $u_k \equiv u_{k-1}, l_k \equiv x_{k-1}$ (shift the lower bound up)
(And if $f(x_{k-1}) = x_{k-1}$ we are done, so we can ignore this case).
This part is kind of like the intermediate value theorem, where we keep narrowing the search for our target.
The set $f([l_k, u_k])$ is in the square  $[l_k, u_k]\times[l_k, u_k]$. (Using induction and $f$ is increasing. It's kind of clear if you draw a picture.)
For any $k$, the square $[l_k, u_k]\times[l_k, u_k]$ intersects the diagonal line $\{ (x, x )\}$ since it contains $(l_k, l_k)$, so if the sequence of nested squares converges, it converges to a point on the diagonal.
But it does converge, since $[l_k, u_k]$ converges to some singleton set $\{ c \}$. 
Putting it all together, $\forall k, f(c) \in f([l_k, u_k]) \subset [l_k, u_k]\times[l_k, u_k]$. 
$ f([c]) \in \cap_k [l_k, u_k]\times[l_k, u_k]$. 
$ f([c]) \in \{ (c, c) \}$. 
$ f(c) = c $. 
