Real Addition is Computably Uniformly Continuous

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Theorem

Let $f : \R^2 \to \R$ be defined as:

$\map f {x, y} = x + y$

Then $f$ is computably uniformly continuous.


Proof

We will define $d : \N \to \N$ as:

$\map d n = 2 n + 1$

which is primitive recursive by:

and thus total recursive by:


Let $x_1, y_1, x_2, y_2 \in \R$ and $n \in \N$ be arbitrary, and suppose:

$\norm {\tuple {x_1, y_1} - \tuple {x_2, y_2}} < \dfrac 1 {\map d n + 1} = \dfrac 1 {2 n + 2}$

We have:

\(\ds \size {x_1 - x_2} + \size {y_1 - y_2}\) \(=\) \(\ds \norm {\tuple {x_1 - x_2, 0} } + \norm {\tuple {0, y_1 - y_2} }\)
\(\ds \) \(\le\) \(\ds \norm {\tuple {x_1 - x_2, y_1 - y_2} }\)
\(\ds \) \(=\) \(\ds \norm {\tuple {x_1, y_1} - \tuple {x_2, y_2} }\) Triangle Inequality for Vectors in Euclidean Space

and thus:

$\size {x_1 - x_2} \le \norm {\tuple {x_1, y_1} - \tuple {x_2, y_2}} < \dfrac 1 {2 n + 2}$
$\size {y_1 - y_2} \le \norm {\tuple {x_1, y_1} - \tuple {x_2, y_2}} < \dfrac 1 {2 n + 2}$


Hence:

\(\ds \size {\map f {x_1, y_1} - \map f {x_2, y_2} }\) \(=\) \(\ds \size {x_1 + y_1 - x_2 - y_2}\) Definition of $f$
\(\ds \) \(=\) \(\ds \size {\paren {x_1 - x_2} + \paren {y_1 - y_2} }\)
\(\ds \) \(\le\) \(\ds \size {x_1 - x_2} + \size {y_1 - y_2}\) Triangle Inequality for Real Numbers
\(\ds \) \(<\) \(\ds \dfrac 1 {2 n + 2} + \dfrac 1 {2 n + 2}\) Inequality above
\(\ds \) \(=\) \(\ds \dfrac 1 {n + 1}\)

$\blacksquare$

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