(a) Recall that the set $K$ is closed in $\mathbb{R}_{K}$ (but not in $\mathbb{R}$ ). Therefore, every one-point set $\{[1]\}$ and $\{x\}$ , $x\notin K$ , in $Y$ is closed, and it is a $T_{1}$ -topological space. At the same time, there is no neighborhood of $0$ in $\mathbb{R}$ that does not contain points in $K$ . Further, every neighborhood of $0$ in $\mathbb{R}_{K}$ is of the form $U$ or $U-K$ where $U$ is open in $\mathbb{R}$ , and, hence, contains a point $x\in K$ . Then, $U$ intersects every neighborhood $V$ of $x$ in $\mathbb{R}_{K}$ . Overall, every neighborhood of $0$ in $Y$ that does not contain $[1]$ is of the form $U-K$ , and every neighborhood of $[1]$ is the union of some neighborhoods of all points of $K$ in $\mathbb{R}$ , and the two must intersect as we said.

(b) $Y$ is not Hausdorff, therefore, the diagonal $\Delta$ of $Y\times Y$ is not closed (by (a), every neighborhood of $0\times[1]$ contains some $x\times x$ : see §17, Hausdorff Spaces). At the same time, $$\begin{eqnarray*} (p\times p)^{-1}(\Delta) & = & \{x\times y\in\mathbb{R}_{K}\times\mathbb{R}_{K}|x=y\mbox{ or }x,y\in K\}\\ & = & \Delta_{K}\cup(K\times K) \end{eqnarray*}$$ is closed as $\mathbb{R}_{K}$ is Hausdorff (the diagonal $\Delta_{K}$ is closed) and $K$ is closed in $\mathbb{R}_{K}$ .

Note that $p$ is not an open quotient map, in particular, $B=p((\tfrac{1}{2},2))=(\tfrac{1}{2},1)\cup(1,2)\cup\{[1]\}$ is not open as $[1]$ has no neighborhood contained in $B$ .