I have such task:
Let $(L, \le)$ be a complete lattice and $f: L \to L$ a monotonic function. Proof that there exists $x \in L$ which is a fixed point of $f$ and is the smallest $y$ that $f(y) \le y$.
I believe it has much to do with the Knaster–Tarski theorem. I think that by proving the Knaster–Tarski theorem I show that there in fact exists such element. But I have problems understanding the theorem's proof. Or perhaps there's a better and easier way to prove that.
If anyone could explain that in simple words I would be very thankful.
Indeed, this is the Knaster-Tarski Fixed Point Theorem.
Forget about the fixed-point part for a bit. Look at the set $$ A:=\{y\in L: f(y)\le y\} $$ We'd like to look at the minimum of $A$, but this need not exist in a general lattice. So, we take the next best thing, namely $x:=\inf (A)=\bigwedge A$.
Fix $y\in A$. We know that $x\le y$, so $f(x)\le f(y)\le y$. This shows that $f(x)$ is a lower bound for $A$, hence that $f(x)\le x$. Applying $f$ again, $f(f(x))\le f(x)$, so $f(x)\in A$. Since $x$ is a lower bound for $A$, $x\le f(x)$. Therefore, $f(x)=x$.
We've argued that $x$ is a fixed point of $f$. Moreover, we've shown that $x=\bigwedge A\in A$, so that we actually have $x=\min(A)$, which is what we wanted to show.