Projection on Real Euclidean Plane is not Closed Mapping

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Theorem

Let $\left({\R^2, d}\right)$ be the real Euclidean plane.

Let $\rho: \R^2 \to \R$ be the first projection on $\R^2$ defined as:

$\forall \left({x, y}\right) \in \R^2: \rho \left({x, y}\right) = x$


Then $\rho$ is not a closed mapping.


The same applies with the second projection on $\R^2$.


Proof

Consider the set $A \subseteq R^2$ of all points defined as:

$A := \left\{{\left({x, y}\right) \in \R^2: x y \ge 1}\right\}$

By Subset of Euclidean Plane whose Product of Coordinates are Greater Than or Equal to 1 is Closed:

$A$ is a closed set in $\left({\R^2, d}\right)$.

By inspection, it can be seen that the image of $A$ under $\rho$ is:

$\rho \left[{A}\right] = \left({\gets \,.\,.\, 0}\right) \cup \left({0 \,.\,.\, \to}\right)$

which by Union of Open Sets of Metric Space is Open is open.

Hence the result by definition of closed mapping.

$\blacksquare$


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