(a) The sets cover $X$ ((iii) and (iv) is already enough). Now, the non-empty intersections: (i) and (i) is of type (i), (i) and (ii) is (i), (i) and (iii) is (i), (i) and (iv) is (i), (ii) and (ii) is (ii), (ii) and (iii) is (ii), (ii) and (iv) is (ii), (iii) and (iii) is (iii), (iii) and (iv) contains (ii), (iv) and (iv) is (iv) (note, once again, the intersection is assumed to be non-empty).(b) $f^{-1}(c)=\cap_n f^{-1}((c-1/n,c+1/n))$ , $f^{-1}(f(p_{n,k}))$ and its intersection with $C_{n,k}$ is a $G_\delta$ set containing $p_{n,k}$ . Therefore, there are only countably many points (given by the intersection of (ii)-type neighborhoods) in $S_{n,k}$ , the set of $p\in C_{n,k}$ such that $f(p)\neq f(p_{n,k})$ . If $y=d$ does not intersect any of $S_{n,k}$ (countably many points) then for every $x\times d\in C_{n,k}$ , $f(x\times d)=f(p_{n,k})$ . Since $f$ is continuous and every neighborhood of a point in $L_m$ contains the intersection of some horizontal interval containing the point with $X$ , $f((n-1)\times d)=f((n+1)\times d)$ . Now, every open set containing $a$ , contains the whole bunch of sets $L_m$ , and the same is true for open sets containing $b$ . Therefore, both points a limit points of $\{m\times d\}$ and $f(a)=f(b)$ .(c) It is not completely regular because $a$ and $b$ cannot be separated (and we will show that one-point sets are closed). Now, it is a $T_1$ space: consider points $a$ , $b$ , two points in $L_m$ and a point in $L_n$ and two points in $C_{n,k}$ and a point in $C_{n',k'}$ . For each point in the list we can easily find a neighborhood that does not contain another point in the list. Now, we show it is regular: every neighborhood of a point contains the closure of another neighborhood.Let $L^-_m$ and $L^+_m$ denote all points to the left (including $a$ ) and to the right (including $b$ ) of $L_m$ . Then $L_m\cup L^-_m$ , $L_m\cup L^+_m$ and $L_m$ are closed. Therefore, any basis neighborhood of $a$ (or $b$ ) contains the closure of another neighborhood.$C_{n,k}$ is closed (for any point not in it find a neighborhood that does not intersect $C_{n,k}$ ). If a point is in $C_{n,k}-p_{n,k}$ then the one-point set is open (the intersection of a small enough horizontal interval and $C_{n,k}$ ) and closed. Every basis neighborhood of $p_{n,k}$ contains all but finite number of points in $C_{n,k}$ , but it is closed as well.If a point is in $L_m$ then its basis neighborhood is either $L^-_{m'}$ for some $m'>m$ or $L^+_{m"}$ for some $m"<m$ or the intersection of a horizontal open interval with $X$ . The first two types of neighborhoods contain a neighborhood of the third type. But the third type neighborhood (small enough so that they do not contain other limit points) is closed as well (the rest of the space is the union of horizontal intersections, some neighborhoods of $a$ and $b$ , and sets $C_{n,k}$ minus finite number of points).