Reals are Close Packed

Theorem

Let $a, b \in \R$ such that $a < b$.

Then:

$\exists c \in \R: a < c < b$

That is, the set of real numbers is close packed.

Proof

We can express $a$ and $b$ as $a = \dfrac a 1, b = \dfrac b 1$.

Thus from Mediant is Between:

$\dfrac a 1 < \dfrac {a + b} {1 + 1} < \dfrac b 1$

Hence $x = \dfrac {a + b} 2$ is an element of $\R$ between $a$ and $b$.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 1.8 \ (6)$