# Upper Semi-Continuous Collections

#### joypav

##### Active member
$G$ is an upper semi-continuous collection of sets filling up the Hausdorff space $X$.

Problem:
No point of $X$ belongs to two elements of $G$.

Proof:

What we know:
$G$ is upper semi-continuous $\implies$ if $g \in G$ and $U$ is an open set containing $g$, then there is an open set $V$ containing $g$ such that each member of $G$ which intersects $V$ lies in $U$

$G$ "fills up" $X \implies X=\cup G$

$X$ is Hausdorff $\implies \forall x_1, x_2 \in X, \exists U_1, U_2$ open such that $x_1 \in U_1, x_2 \in U_2, U_1 \cap U_2=\emptyset$

Here's my questions...
Why can't an element of $X$ be in two $g$'s, say $g_1$ and $g_2$ of the collection $G$? I know that's the question I am being asked, but what I mean is, say $g_1 \subset g_2$. Then $x$ could be in both and the same open sets could satisfy the requirements for upper semi-continuity for both.

#### joypav

##### Active member
Aha! I was thinking too deep about it... we only need to show the existence of ONE open set that contains one and not the other. I will post my proof for completeness.

Proof:
BWOC, assume $\exists g_1, g_2 \in G$ such that $g_1 \cap g_2 \neq \emptyset$.
Consider $y \in g_2$ such that $y \notin g_1$. (WLOG, assume $g_1 \not\subset g_2$. If it was we could just switch around $g_1$ and $g_2$ in the proof.)

$X$ Hausdorff $\implies$ for each $x \in g_1 \exists U_x, V_x$ open in $X$ such that $x \in U_x, y \in V_x, U_x \cap V_x = \emptyset$.

Let $U = \cup_{x \in g_1} U_x$.
Then, obviously, $g_1 \subset U$, where $U$ is and open set of $X$. (it is the union of open sets)
$G$ is upper semi-continuous $\implies \exists V$ open in $X$ such that $g_1 \subset V$ and if $g' \in G$ with $g' \cap V \neq \emptyset$ then $g' \subset U$

Claim: $g_2 \cap V \neq \emptyset$
By assumption, $g_1 \cap g_2 \neq \emptyset$ and we know $g_1 \subset V \implies V \cap g_2 \neq \emptyset$
$\implies$ our Claim is true.

So we've shown that $g_2$ does in fact intersect $V$ (hoorah!).
$\implies g_2 \subset U$ (by definition of upper semi-continuity)
Go back to the element we considered in $g_2$. We know $y \in g_2$.
Then, $y \in g_2 \implies y \in \cup_{x \in g_1} U_x \implies y \in U_x$ for some $x \in g_1$.
But this is a contradiction, because $y \in V_x$ for every $x$ and $U_x \cap V_x = \emptyset$.