Real and Imaginary Part Projections are Continuous

From ProofWiki
Jump to navigation Jump to search

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 $\cmod {w - z} < \delta$:

\(\ds \cmod {\map \Re w - \map \Re z}\) \(=\) \(\ds \cmod {\map \Re w +i\map\Im w- \map \Re z-i\map\Im z+i\map\Im z-i\map\Im w}\)
\(\ds \) \(\le\) \(\ds \cmod {\map \Re w +i\map\Im w- \map \Re z - i \map \Im z}+\cmod{i\map\Im z-i\map\Im w}\) Triangle Inequality for Complex Numbers
\(\ds \) \(\le\) \(\ds \cmod {\map \Re w +i\map\Im w- \map \Re z - i \map \Im z}\) modulus is always non-negative
\(\ds \) \(=\) \(\ds \cmod{w-z}\)
\(\ds \) \(<\) \(\ds \delta\)
\(\ds \) \(=\) \(\ds \epsilon\)

and

\(\ds \cmod {\map \Im w - \map \Im z}\) \(=\) \(\ds \cmod i \cmod {\map \Im w - \map \Im z}\) $\cmod i=1$
\(\ds \) \(=\) \(\ds \cmod {i\map \Im w - i\map \Im z}\) Complex Modulus of Product of Complex Numbers
\(\ds \) \(=\) \(\ds \cmod {\map \Re w +i\map\Im w- \map \Re z-i\map\Im z+\map\Re z-\map\Re w}\)
\(\ds \) \(\le\) \(\ds \cmod {\map \Re w +i\map\Im w- \map \Re z - i \map \Im z}+\cmod{\map\Re z-\map\Re w}\) Triangle Inequality for Complex Numbers
\(\ds \) \(\le\) \(\ds \cmod {\map \Re w +i\map\Im w- \map \Re z - i \map \Im z}\) modulus is always non-negative
\(\ds \) \(=\) \(\ds \cmod{w-z}\)
\(\ds \) \(<\) \(\ds \delta\)
\(\ds \) \(=\) \(\ds \epsilon\)

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.