Blackwell's Sufficiency Conditions for a Contraction: I wondered if anyone could throw some intuition on Blackwell's sufficiency conditions:
(1) Monotononicty;
(2) Discounting;
Would be genuinely appreciated to hear both terms expressed in plain English.
Let $X$ be a nonempty set, $B(X)$ be the space of bounded functions from $X$ to $\mathbb{R}$, and $\|\cdot\|_\infty$ the sup-norm on $B(X)$. Write $f\leq g$ for $f,g\in B(X)$ if $f(x)\leq g(x)$ for all $x\in X$. Let $\mathbf{1}\in B(x)$ be the function with constant value $1$.
Here is the result of Blackwell:

Theorem: Let $T:B(X)\to B(X)$ be a function such that for some $\beta\in(0,1)$

Whenever $f,g\in B(X)$ satisfy $f\leq g$, then $T(f)\leq T(g)$ (monotonicity).
For all $f\in B(X)$ and $\alpha\geq 0$ one has $T(f+\alpha\mathbf{1})\leq T(f)+\alpha\beta\mathbf{1}$ (discounting).

Then, $\|T(f)-T(g)\|_\infty\leq \beta\|f-g\|_\infty$ for all $f,g\in B(X)$.

The following argument should make things a bit more transparent. Take $\alpha\geq 0$ large enough that both $f\leq g+\alpha\mathbf{1}$ and $g\leq f+\alpha\mathbf{1}$ holds. Since both $f$ and $g$ are bounded, this is certainly possible.
By monotonicity,
$$T(f)\leq T\big(g+\alpha\mathbf{1}\big).$$
By discounting,
$$T(f)\leq T(g)+\beta\alpha\mathbf{1}.$$
Therefore, $$T(f)-T(g)\leq \beta\alpha\mathbf{1}.$$
By switching $f$ and $g$, we get
$$T(g)-T(f)\leq \beta\alpha\mathbf{1}.$$
It follows that
$$|T(g)(x)-T(f)(x)|\leq \beta\alpha$$
for all $x\in X$,
which implies
$$\|T(f)-T(g)\|_\infty\leq \beta\alpha.$$
To finish the proof, simply note that we can choose $\alpha=\|f-g\|_\infty$.
Here, monotonicity guarantees that $T$ preserves the pointwise ordering of functions, and discounting guarantees that if we increase every coordinate of a function by the same amount, the resulting value of $T$ increases at most $\beta$ times this amount.
We want to use discounting even for two functions in which neither equals the other plus a constant amount. Monotonicity guarantees us that if one function is bounded above by another function plus a constant value, a similar bound can be used for the transformed functions under $T$. This makes discounting much more powerful.