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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {n + 1}\) |
$\blacksquare$