Continuous Monster

Dirichlet Monster Function (reverse of Monster 3): $latex D(x) = \begin{cases} 1/b, & \text{if }x=a/b\in\mathbb{Q}\text{ rational} \\ 0, & \text{if }x\text{ irrational} \end{cases} $ Weirstrass proved by 'ε-δ' that D(x) is continuous at any irrational x. Proof: Take any irrational xo =√2 Let x= 13/10 (the smallest rational x=a/b nearest to xo) |x-xo|=|13/10 - √2|= 0.114 < δ Take δ =0.2 => |D(x) - D(xo)|= |D(13/10)-D(√2)| = 1/10 - 0 = 0.1 < ε=δ=0.2 For all ε, there is δ=ε such that |x-xo|< δ => |D(x) - D(xo)|< ε

Hence,
D(x) is continuous at irrational xo.

[QED]
---------------------------

Note:

Monster 3:

$latex
f(x) =
\begin{cases}
1, & \text{if }x \text{ is irrational} \\
0, & \text{if }x\text{ rational}
\end{cases}
$