Question: Suppose $(X, < )$ is a totally ordered set. Then

$$ (a, \infty) = \bigcup\_{x > a} [x, \infty) $$

for any $a \in X$

### Attempt:

Let $y \in \bigcup\_{x > a} [x, \infty) $. So there is some $x > a $ so that $y \geq x $ and so $y > a $. Hence, $x \in (a, \infty) $.

As for the other direction, I am kind of stuck. Would be wise to prove to contrapositive instead?

Answer: 
Notice if $d, e$ are random variables uniform over $[0,1]$, so do $\tilde{d} \stackrel{def}{=} 1 - d\;$ and $\;\tilde{e}\stackrel{def}{=} 1 - e$.  

We have
$$a + b + c > d + e \quad\iff\quad a + b + c + \tilde{d} + \tilde{e} > 2$$
For any $p = (x\_1,x\_2,x\_3,x\_4,x\_5) \in \mathbb{R}^5$, let $\;\ell(p) = x\_1 + x\_2 + x\_3 + x\_4 + x\_5$.  

The probability $\mathcal{P}$ we seek is equal to:

$$\mathcal{P} = \verb/Volume/\big\{\; p \in [0,1]^5 : \ell(p) > 2 \;\big\}
= 1 - \verb/Volume/\big\{\; p \in [0,1]^5 : \ell(p) \le 2 \;\big\}\tag{\*1}$$

Let $e\_1, e\_2, \ldots, e\_5$ be the standard bases of $\mathbb{R}^5$ and $\Delta\_p(\lambda)$ be the simplex 
$$\big\{\; q \in \mathbb{R}^5 : q\_i \ge p\_i, i = 1\ldots5\;\text{ and }\; \ell(q-p) \le \lambda \;\big\}$$

The polytope on the right of $(\*1)$ can be constructed by
cutting 5 small simplices based at $e\_i$ from the simplex $\Delta\_0(2)$. More precisely,
up to a set of measure zero, we have:
$$\big\{\; p \in [0,1]^5 : \ell(p) \le 2 \;\big\}
= [0,1]^5 \cap \Delta\_0(2)
\approx \Delta\_0(2) \setminus \left( \bigcup\_{i=1}^5 \Delta\_{e\_i}(1) \right)$$
This leads to$\color{blue}{^{[1]}}$

$$\mathcal{P}
= 1 - \frac{2^5}{5!} + 5 \left(\frac{1^5}{5!}\right)
= 1 - \frac{32 - 5}{120} = \frac{31}{40}$$

**Notes**

* $\color{blue}{[1]}$ In general, in any dimension, the volume of the simplex $\Delta\_0(\lambda)$ inside the unit hyper-cube is given by following formula:
$$\verb/Volume/\big( [0,1]^n \cap \Delta\_0(\lambda) \big) = \sum\_{k=0}^{\lfloor\lambda\rfloor} (-1)^k \binom{n}{k} \frac{(\lambda-k)^n}{n!}$$
One can prove this formula using [inclusion exclusion principle](http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle). Look at my [answer](https://math.stackexchange.com/a/455711/59379) for a similar question in $\mathbb{R}^3$,
it has more explanation of the underlying ideas.
$$\int\_0^1 da \int\_0^1 db \int\_0^1 dc \int\_0^{{\rm min}(1,a+b+c)} dd \int\_0^{{\rm min}(1,a+b+c-d)} de = \frac{31}{40}=0.775$$