I am trying to understand an example for a space which is not Hausdorff.
I do not really see, why (Q1******) and (Q2******)marked underneath hold.
To show: $Y$ (see definition underneath) is not Hausdorff.
$X = [-1,1] \times \{0, 1 \} \subset \mathbb{R}^2$, with induced topology of $\mathbb{R}^2$ ($X$ is Hausdorff).
$Y = X /_{\sim}$, $\sim$ induced by: $(t,0) \sim (t,1) \ \forall t \in [-1,1] \setminus \{ 0 \}$
Let $P =(0,1), Q = (0,0)$. We will show: $P$ and $Q$ cannot be separated.
Let $U$ be open in $Y$, $P \in U$ and $V$ open in $Y$, $Q \in V$.
Let $p: X \to Y$ be the projection. Then
$p^{-1} (U) = U' \subset X$ open, $p^{-1} (V) = V' \subset X$ open.
Furthermore: $(0,1) \in U'$, $(0,0) \in V'$ and $U', V'$ open sets.
Obviously: $(- \epsilon, \epsilon) \times \{ 1 \} \subset U'$ since $(0,1) \in U'$.
$p( (-\epsilon, \epsilon)) \supset ((-\epsilon,0) \cup (0,\epsilon))
\cup \{ P \} $ (Q1******)
$\Rightarrow U' \supset ((- \epsilon, \epsilon) \times \{ 1 \} ) \cup
((- \epsilon, 0) \cup (0, \epsilon)) \times \{ 0 \} )$ (Q2******)
$\Rightarrow U' \supset ((- \epsilon, 0) \cup (0, \epsilon)) \times \{ 0,1 \} $
Similarly, $V' \supset ((- \epsilon', 0) \cup (0, \epsilon')) \times \{ 0,1 \} $
$\Rightarrow U' \cap V' \neq \emptyset$
$\Rightarrow U \cap V \neq \emptyset$.
Best Answer
There's indeed a mild ambiguity in the notation.
To visualize $Y$, it's almost a single ordinary line segment, but its midpoint is taken as two different points (the two parallel line segments of $X$ are collapsed into one except for their midpoints). $$\_\_\_\_\_{}_:\_\_\_\_\_$$
Now, $(-\epsilon,0)\cup(0,\epsilon)$ in Q1 refers to a set of points on the two single line segment parts of $Y$ (each point with double cover in $X$), specifically to the set $$\{[(a,0)]_\sim\,\mid\,-\epsilon<a<\epsilon,\,a\ne 0\}\ =\ \{[(a,1)]_\sim\,\mid\,-\epsilon<a<\epsilon,\,a\ne 0\}\,,$$ and indeed its preimage is $(-\epsilon,0)\times\{0,1\}\,\cup\,(0,\epsilon)\times\{0,1\}$.
If you add $P=[(0,1)]_\sim$ to this set, then it only adds the point $(0,1)$ in the preimage, as $P$ is not doubly covered.
The same goes for $Q$.