Real and Imaginary Part Projections are Continuous

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Theorem

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

$\forall z \in \C: x \left({z}\right) = \operatorname{Re} \left({z}\right)$
$\forall z \in \C: y \left({z}\right) = \operatorname{Im} \left({z}\right)$


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 $\left\vert{w - z}\right\vert < \delta$:

\(\displaystyle \left\vert{\operatorname{Re} \left({w}\right) - \operatorname{Re} \left({z}\right)}\right\vert + \left\vert{\operatorname{Im} \left({w}\right) - \operatorname{Im} \left({z}\right) }\right\vert\) \(=\) \(\displaystyle \left\vert{\operatorname{Re} \left({w}\right) - \operatorname{Re} \left({z}\right)}\right\vert + \left\vert{i \left({\operatorname{Im} \left({w}\right) - \operatorname{Im} \left({z}\right)}\right)}\right\vert\) as $\left\vert{i}\right\vert = 1$
\(\displaystyle \) \(\le\) \(\displaystyle \left\vert{\operatorname{Re} \left({w}\right) - \operatorname{Re} \left({z}\right) + i \operatorname{Im} \left({w}\right) - i \operatorname{Im} \left({z}\right)}\right\vert\) Triangle Inequality for Complex Numbers
\(\displaystyle \) \(=\) \(\displaystyle \left\vert{w - z}\right\vert\)
\(\displaystyle \) \(<\) \(\displaystyle \epsilon\)



This equation shows that $\left\vert{x \left({w}\right) - x \left({z}\right)}\right\vert < \epsilon$, and $\left\vert{y \left({w}\right) - y \left({z}\right) }\right\vert < \epsilon$.

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 Topological Product.

Suppose we let $z = \operatorname{Re} \left({z}\right) + i \operatorname{Im} \left({z}\right)$ be a complex number.

We can now identify the complex number $z$ with the ordered pair $\left({\operatorname{Re} \left({z}\right), \operatorname{Im} \left({z}\right)}\right) \in \R^2$, 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.