Image under Projection from Closed Set of Product Topology is not necessarily Closed

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.

Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.

Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors.


Let $K \subseteq S_1 \times S_2$ be closed in $T$.

Then $\pr_1 \sqbrk K$ is not necessarily closed in $T_1$.


Proof

Proof by Counterexample:

Let $K = \set {\tuple {x, y} \in \R^2: x \ge 0, y \ge \dfrac 1 x}$

Then:

$\pr_1 \sqbrk K = \openint 0 \to$

$K$ is closed in $T$, as follows:

Consider the mapping $f: \R^2 \to \R$ defined as:

$\map f {x, y} = x y$

which is continuous on $\R^2$ (see Preimages of $\map f {x, y} = \tuple {x^2 + y^2, x y}$: Continuity).

Then we note that: $K = f^{-1} \hointr 1 \to \cap p_1^{-1} \hointr 0 \to$

and the result follows.

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Image_under_Projection_from_Closed_Set_of_Product_Topology_is_not_necessarily_Closed&oldid=505672"