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[SOLVED] Upper Semi-Continuous Collection

joypav

Active member
Mar 21, 2017
151
Problem:
Suppose that $X$ and $Y$ are each compact sets. Consider $X \times Y$.
Let $g_x = \left\{x\right\} \times Y$ and $G = \left\{g_x : x \in X\right\}$.
Show that if $g_x \in G$ and $U \subset X \times Y$ open with $g_x \subset U$, then $\exists V$ open in $X \times Y$ such that:
(i) $g_x \subset V$
(ii) for any $g_x' \in G$, if $g_x' \cap V \neq \emptyset$ then $g_x' \subset U$
(meaning, $G$ is an upper semi-continuous collection)

If anyone has any hints to give I would appreciate it!

We are looking at vertical strips.
I considered $g_x \in G$, some $x \in X$, and an open set $U$ containing $g_x$. Of course I need to construct a $V$ that satisfies these conditions. Any hints? I thought maybe to use compactness, as $X \times Y$ is compact and every open cover has a finite subcover.
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,890
Hi joypav ,

If $U$ is open in $X\times Y$ and $g_x \subset U$, then to each $y\in Y$ correspond open neighborhoods $O_y\ni x$ and $V_y\ni y$ such that $O_y \times V_y \subset U$. Note $Y$ is covered by the open sets $V_y$, so by compactness there are $y_1,\ldots, y_n\in Y$ such that $Y = V_{y_1}\cup \cdots \cup V_{y_n}$. Set $O = O_{y_1}\cap \cdots \cap O_{y_n}$. Then $O$ is an open neighborhood $x$ such that $O \times Y \subset U$. Set $V = O \times Y$.
 

joypav

Active member
Mar 21, 2017
151
Hi joypav ,

If $U$ is open in $X\times Y$ and $g_x \subset U$, then to each $y\in Y$ correspond open neighborhoods $O_y\ni x$ and $V_y\ni y$ such that $O_y \times V_y \subset U$. Note $Y$ is covered by the open sets $V_y$, so by compactness there are $y_1,\ldots, y_n\in Y$ such that $Y = V_{y_1}\cup \cdots \cup V_{y_n}$. Set $O = O_{y_1}\cap \cdots \cap O_{y_n}$. Then $O$ is an open neighborhood $x$ such that $O \times Y \subset U$. Set $V = O \times Y$.
Thanks!

Looking at the decomposition space, $X \times Y / G$, as this would be the space where each $g_x$ is a point of the space, would it be reasonable to expect that it be homeomorphic to $X$ itself?
 

Euge

MHB Global Moderator
Staff member
Jun 20, 2014
1,890
Yes. For consider the projection $\pi_X : X \times Y \to X$ sending $(x,y)$ in $X\times Y$ to $x$. Since $\pi_X(g_x) = \{x\}$ for all $x\in X$, the universal property of quotient spaces admits a continuous mapping $X\times Y/G \to X$ sending an equivalence class $[x,y]$ to $x$. Fix $y_0$ in $G$ and define a mapping $X \to X\times Y/G$ sending $x$ to $[x,y_0]$. It is a continuous inverse of the map $X \times Y/G \to X$, so $X\times Y/G$ is homeomorphic to $X$.