Show that the projection is, in general, not closed

Question

I have $(X,d) \text{ and } (Y,d')$ two metric spaces. I define the metric space $(X \times Y, d_{\infty})$ and the projections $\pi_1:X \times Y \rightarrow X$, $\pi_2:X \times Y \rightarrow Y$. I've already shown that these functions are continous and open, but can't find an example for where it's not closed

Answer 1

A classical example is $X=Y=\mathbb R$. The sup metric on $\mathbb R^2$ induces the Euclidean topology.

Let now $C=\{xy=1\}$. It is closed in $\mathbb R^2$ but it's project on the X-axis is $\{x\neq0\}$, which is not closed.

Answer 2

Let $X=Y=\mathbb{R}$ and let $C=\{(x,y)\in (0,\infty)\times (0,\infty)\;|y\geq 1/x \}$

Then, $C$ is closed. Indeed, if $(x_n,y_n)\in C$ with limit $(x,y)$, then either $x\neq 0$, in which case $(x,y)\in C$ by continuity of $x\mapsto \frac{1}{x}$, or $x=0$, in which case the $y_n$ sequence is unbounded and thus, not convergent. We conclude that $C$ is closed.

However, $\pi_1(C)=(0,\infty),$ which clearly isn't closed.

Answer 3

The "standard" example:

Take $(X,d)=(\Bbb R, d_e)=(Y,d')$ where $d_e(x,y)=|x-y|$ the usual Euclidean metric. $d_\infty$ is equivalent to the usual metric on $\Bbb R^2$.

Show that $A=\{(x,y) \in \Bbb R^2: xy=1\}$ is a closed set (the inverse image of $\{1\}$ under a continuous map) and $\pi_1[A]=\pi_2[A]= \Bbb R\setminus \{0\}$ which are not closed in the factor spaces. So $\pi_1$ nor $\pi_2$ are closed.

If $X$ is compact, $\pi_2$ is a closed map, and if $Y$ is, so is $\pi_1$ (projections along a compact space are closed). So for such an example we need non-compact spaces.

Answer 4

The set $\{(x,y)\in\mathbb R^2\mid xy\geq1\}$ is a closed subset of $\mathbb R^2$.

However it is projected to $\mathbb R-\{0\}$ which is not closed in $\mathbb R$.