1. **Problem Statement:** Show that $(X, \tau)$ with $X = \{a,b,c\}$ and $\tau = \{\emptyset, X, \{a\}, \{b\}, \{a,b\}\}$ is a topological space and find the closure of the set $\{a\}$. 2. **Definition of a Topological Space:** A topological space is a set $X$ together with a collection $\tau$ of subsets of $X$ satisfying: - $\emptyset$ and $X$ are in $\tau$. - The union of any collection of sets in $\tau$ is also in $\tau$. - The finite intersection of sets in $\tau$ is also in $\tau$. 3. **Check the Axioms:** - $\emptyset$ and $X$ are in $\tau$ by definition. - Unions: - $\{a\} \cup \{b\} = \{a,b\} \in \tau$. - Any union of sets in $\tau$ is either $\emptyset$, $\{a\}$, $\{b\}$, $\{a,b\}$, or $X$, all in $\tau$. - Finite intersections: - $\{a\} \cap \{b\} = \emptyset \in \tau$. - $\{a\} \cap \{a,b\} = \{a\} \in \tau$. - All finite intersections remain in $\tau$. 4. **Conclusion:** $\tau$ satisfies the axioms of a topology, so $(X, \tau)$ is a topological space. 5. **Find the Closure of $\{a\}$:** - The closure $\overline{\{a\}}$ is the smallest closed set containing $\{a\}$. - Closed sets are complements of open sets in $\tau$: - $\emptyset^c = X$ (closed) - $X^c = \emptyset$ (closed) - $\{a\}^c = \{b,c\}$ (not in $\tau$, so not open, so $\{a\}$ is not closed) - $\{b\}^c = \{a,c\}$ (not open) - $\{a,b\}^c = \{c\}$ (not open) - The closed sets are $\emptyset$ and $X$ only. - Since $\{a\} \subseteq X$ and $X$ is closed, $\overline{\{a\}} = X$. **Final answer:** $(X, \tau)$ is a topological space and the closure of $\{a\}$ is $X$.