This question was asked in my Topology quiz (now over) and I was unable to prove it
Let $A=\{1/n : n \in \mathbb{N}\}$, and let $$B' = \{B\in \mathcal P(\mathbb{R}):(B\text{ is an open interval s.t. } 0\notin B)\vee(B= (-x,x)\setminus A\mbox{ for a }x>0)\}$$Then B' is a basis for a topology $T$ on $\mathbb{R}$, and the space ($\mathbb{R}, T$) is a Hausdorff space. Let $D$ be the decomposition of $\mathbb{R}$ whose members are $A$ and $\{x\}$ for all $x\notin A$.
If $U$ is the quotient topology on $D$ induced by the natural map $p: \mathbb{R} \rightarrow D$, then prove that $p$ is closed.
This might be useful: Let $(X,T)$ be a topological space, and let $D$ be a partition of $X$. Define a function $p: X\rightarrow D$ as follows: For each $x\in X$, $p(x)$ is the member of $D$ that contains $x$. If $U$ is the quotient topology on $D$ induced by $p$, then $(D,U)$ is a quotient map. The function $p$ is called natural map of $X$ onto $D$.
I am really sorry but I would not be able to provide attempt for this question as I was unable to solve it during quiz and even now while trying at home nothing is coming clear. I am not able to see a reason why this map is closed.
Definition: A decomposition $\mathscr{D}$ is said to be upper semicontinuous iff for each $ F\in\mathscr{D}$ and each open set $U$ in $X$ containing $F$, there is some saturated open set $V$ in $X$ with $F\subseteq V\subseteq U$.
Theorem: The natural map $P$ associated with a decomposition space $\mathscr{D}$ of $X$ is closed iff $\mathscr{D}$ is upper semicontinuous.
We show that the decomposition is upper semicontinuous. Suppose $F\in\mathscr{D}$ and $U$ is an open set containing $F$. First, let $F = \{x\}$ where $x\notin A\cup\{0\}$. Then one can find an open interval $V\subseteq U$ containing $F$ that does not intersect $A$. For the second case, let $F = \{0\}$. Then there exists $\varepsilon > 0$ such that $V = (-\varepsilon, \varepsilon)-A\subseteq U$. In both these cases it is clear that $V$ is saturated since $V = \bigcup_{y\in V}\{y\}$. Finally, let $F = A$. For each $x\in F$ let $V_{x}\subseteq U$ be an open interval containing $x$. Then $V = \bigcup_{x\in V} V_{x}\subseteq U$ and \begin{equation*} V = \bigg(\bigcup_{x\in V} (V_{x}-A)\bigg)\cup A = \bigg(\bigcup_{x\in V} \bigg(\bigcup_{y\in V_{x}-A}\{y\}\bigg)\bigg)\cup A, \end{equation*} which implies that $V$ is saturated.
More generally, a decomposition space $\mathscr{D}$ of $X$ is said to be finite iff only finitely many elements of $\mathscr{D}$ have more than one point. Show that a finite decomposition space whose elements are closed in $X$ is upper semicontinuous.