Real and Imaginary Part Projections are Continuous

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Theorem

Define the real-valued functions $x, y: \C \to \R$ by:

$\forall z \in \C: \map x z = \map \Re z$
$\forall z \in \C: \map y z = \map \Im z$


Equip $\R$ with the usual Euclidean metric.

Equip $\C$ with the usual Euclidean metric.


Then both $x$ and $y$ are continuous functions.


Proof

Let $z \in \C$, and let $\epsilon \in \R_{>0}$.

Put $\delta = \epsilon$.

For all $w \in \C$ with $\cmod {w - z} < \delta$:

\(\ds \cmod {\map \Re w - \map \Re z} + \cmod {\map \Im w - \map \Im z}\) \(=\) \(\ds \cmod {\map \Re w - \map \Re z} + \cmod {i \paren {\map \Im w - \map \Im z} }\) as $\cmod i = 1$
\(\ds \) \(\le\) \(\ds \cmod {\map \Re w - \map \Re z + i \map \Im w - i \map \Re z}\) Triangle Inequality for Complex Numbers
\(\ds \) \(=\) \(\ds \cmod {w - z}\)
\(\ds \) \(<\) \(\ds \epsilon\)



This equation shows that $\cmod {\map x w - \map x z} < \epsilon$, and $\cmod {\map y w - \map y z} < \epsilon$.

It follows by definition that $x$ and $y$ are both continuous.

$\blacksquare$


Notes


This theorem can be considered a special case of Continuous Mapping to Product Space.

Suppose we let $z = \map \Re z + i \map \Im z$ be a complex number.

We can now identify the complex number $z$ with the ordered pair $\tuple {\map \Re z, \map \Im z} \in \R^2$, where $R^2$ is the Cartesian product $\R \times \R$.

The functions $x$ and $y$ can now be considered as projections on the co-ordinates.