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**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.