# Upper Semi-Continuous Collections, filling up the half-plane

#### joypav

##### Active member
Let $X = [0, ∞) \times R$. Consider the following collections of sets filling up $X$.

(i.) Let $G = \left\{ \left\{ x \right\} \times R : x ≥ 0 \right\}$. Show that $G$ is not upper semi-continuous.

(ii.) Let $G′ = \left\{ \left\{ x \right\} \times [0, 1] : x > 0 \right\}$; let $G = G′ \cup \left\{ \left\{ (x, y) \right\} : x > 0, y \notin [0, 1] \right\} \cup \left\{ \left\{ 0 \right\} \times R \right\}$. Show that $G$ is upper semi-continuous.

(iii.) Let $G′ = \left\{ \left\{ x \right\} \times [0, 1]|x ≥ 0 \right\}$; let $G = G′ \cup \left\{ \left\{(x, y) \right\} : x ≥ 0, y \notin [0, 1] \right\}$.
(So $G$ is $G′$ together with the singletons that are not in any elements of $G′$)
Show that $X/G$ is homeomorphic to $X$.

Proof of (i.):
Let $g_x := \left\{x\right\} \times R$.
Consider $g_0 := \left\{0\right\} \times R \in G$.
Let $U = \left\{ (x, y) : y > 1/x \right\} \cup \left\{ (x, y) : y < 1/x \right\}$.
Then $U$ is open, as it is the union of two open sets.
Note that $U^c = \left\{ (x, y) : xy=1 \right\}$.

Claim: $g_0 \subset U$
$\triangleright$ Let $(x, y) \in g_0 \implies x=0, y \in R \implies xy=0$ for ever $y \implies xy \neq 1$ for every $y \implies (x, y) \notin U^x \implies (x, y) \in U$ //

By way of contradiction, assume that $G$ is upper semi-continuous.
Then, $\exists V$ open such that,
1. $g_0 \subset V$
2. if $g_x \in G$ is such that $g_x \cap V \neq \emptyset$ then $g_x \subset U$

Claim: There is some $g_x$ (different of $g_0$) which intersects $V$.
$\triangleright$ Suppose not. Then, for every $(x, y) \in V, x=0$ (otherwise $V$ would intersect a $g_x$). But $g_0$ must be contained in $V$ by definition, $\implies g_0 = V$, a contradiction. //

So, we have $g_x \neq g_0$ such that $g_x \cap V \neq \emptyset \implies g_x \subset U$ (definition of upper semi-continuous).
Choose $(x, 1/x) \in g_x$. Then, $xy = x(1/x) = 1 \implies g_x \cap U^c \neq \emptyset \implies g_x \not\subset U$, a contradiction.
$\implies G$ is not upper semi-continuous.
This okay?

Thoughts on (ii.):
Checking the individual points...
Consider $(x, y), x>0, y \notin [0,1]$.
Let $U$ be an open set such that $(x, y) \in U$.

Case 1: $U \cap (G' \cup \left\{\left\{0\right\} \times R \right\}) = \emptyset$
Then, just let $V=U$. Need to check this $V$ satisfies 1. and 2.
1. $(x, y) \in U \implies (x, y) \in V$
2. if some $(x', y') \in V \implies (x', y') \in U$

Case 2: $U \cap G' \neq \emptyset$ and $U \cap \left\{\left\{0\right\} \times R \right\} = \emptyset$
Let $F = \left\{g_x \in G' : g_x \cap U \neq \emptyset\right\}$.
Let $V = U - (\cup_{g_x \in F} g_x ) = U \cap (\cup_{g_x \in F} g_x )^c \implies V$ open, as it is the intersection of two open sets.
Then the only things left to intersect $V$ are the individual points, and I can use the same argument as in Case 1 to show $V$ satisfies 1. and 2.
... is this the right idea, or not so much?

I would love some help with (ii.) and (iii.). Though I think figuring out (ii.) would be enough to help me understand (iii.).
I see that for (ii.) I need to check each piece, meaning...
$\cdot$ for the y-axis, if an open set $U$ contains the y-axis I can find a $V$ satisfying u.s.c. criteria
$\cdot$ for each vertical line (going from $y$ is 0 to 1), if an open set $U$ contains one of these lines, then I can find a $V$ satisfying u.s.c. criteria
$\cdot$ for and individual point, if an open set $U$ contains this point then I can find a $V$ satisfying u.s.c. criteria

Also, definitely need help with seeing that the decomposition space is homeomorphic to our space $X$!

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