1. Problem: Find the limit points of given sets in the finite closed topology on integers $(\mathbb{Z}, T)$. (i) Set $A = \{1, 2, 3, \ldots, 10\}$. (ii) Set $E$ consisting of all non-integers. 2. Problem: Given $S \subseteq T$ subsets of a topological space $(X, T)$, prove properties about limit points and density. --- ### Step 1: Understanding finite closed topology on $\mathbb{Z}$ In the finite closed topology, the closed sets are exactly the finite subsets and the whole space. Hence, the open sets are complements of finite sets or empty. ### Step 2: Limit points definition A point $p$ is a limit point of a set $A$ if every open neighborhood of $p$ contains a point of $A$ different from $p$. ### Step 3: Limit points of $A = \{1, 2, \ldots, 10\}$ Since $A$ is finite, and the topology is finite closed, every singleton is closed. Thus, no point outside $A$ can be a limit point because neighborhoods exclude finitely many points. Also, for points in $A$, since neighborhoods can exclude all other points, no point in $A$ is a limit point of $A$. **Conclusion:** $A$ has no limit points. ### Step 4: Limit points of $E$ (all non-integers) Since $E$ contains no integers, and the space is $\mathbb{Z}$, $E$ is empty in this topology. Hence, $E$ has no limit points in $\mathbb{Z}$. ### Step 5: Prove if $p$ is a limit point of $S$ and $S \subseteq T$, then $p$ is a limit point of $T$ Since $p$ is a limit point of $S$, every open neighborhood $U$ of $p$ contains a point of $S$ different from $p$. Because $S \subseteq T$, $U$ also contains a point of $T$ different from $p$. Thus, $p$ is a limit point of $T$. ### Step 6: Deduce from (i) that $S \subseteq T$ This is given as a hypothesis, so no proof needed. ### Step 7: Show if $S$ is dense in $X$, then $T$ is dense in $X$ Density means the closure of $S$ is $X$, i.e., every point of $X$ is either in $S$ or a limit point of $S$. From Step 5, all limit points of $S$ are limit points of $T$. Since $S \subseteq T$, all points of $S$ are in $T$. Hence, closure of $T$ contains closure of $S$, so $T$ is dense in $X$. --- **Final answers:** (i) Limit points of $A$ are $\emptyset$. (ii) Limit points of $E$ are $\emptyset$. (iii) If $p$ is a limit point of $S$ and $S \subseteq T$, then $p$ is a limit point of $T$. (iv) If $S$ is dense in $X$ and $S \subseteq T$, then $T$ is dense in $X$.