What is the algebra generated by the class $\mathcal{E}$ of sets described? What is the algebra generated by the class $\varepsilon$ of sets described?
a) For a fixed subset $E$ of $S$, $\mathcal{E}=\{E\}$ is the class containing $E$ only.
b) For a fixed subset $E$ of $S$, $\mathcal{E}$ is the class of sets of which $E$ is a subset, i.e.: $$\mathcal{E}= \{F∶E \subseteqq F , F \subseteqq S\}$$
c) $\mathcal{E}$ is the class of all sets which contain exactly two points.
 A: As remarked in the comments above, $$\mathcal{A}=\{\emptyset,E,S\setminus E,S\}$$ is an algebra on $S,$ and is the smallest such containing $\mathcal{E}=\{E\}$.

For $\mathcal{E}=\{F\subseteq S:E\subseteq F\},$ note that $\mathcal{E}$ is already closed under arbitrary unions and intersections, so to start getting a handle on the algebra it generates, we need to consider complements first. Letting $\mathcal{A}$ be the algebra on $S$ generated by $\mathcal{E},$ we know that $S\setminus F\in\mathcal{A}$ for all $F\in\mathcal{E},$ which by definition of $\mathcal{E}$ means that every subset of $S\setminus E$ is in $\mathcal{A}.$ That is, every subset of $S$ disjoint from $E$ is an element of $\mathcal A.$ In fact, it is relatively easily shown that if $\mathcal C=\{F\subseteq S:E\cap F=\emptyset\},$ then $\mathcal E\cup\mathcal C$ is an algebra, and so is precisely $\mathcal A.$

For $\mathcal{E}$ the set of two-element subset of $S,$ note that every singleton subset of $S$ is the intersection of two two-element subsets, and every finite subset of $S$ is a finite union of singleton subsets of $S$. Consequently, the algebra $\mathcal{A}$ on $S$ generated by $\mathcal{E}$ will contain every finite subset of $S,$ so will also contain every subset of $S$ having a finite complement. Can you think of any other subsets of $S$ that will be in $\mathcal{A}?$ If not, can you prove that $\mathcal{A}$ contains precisely the finite subsets of $S$ and the subsets of $S$ with finite complements? If you can't think of any others and can't prove that there aren't any others, let me know. Added: As Did pointed out in the comments below, this approach does not work when $S$ has fewer than $3$ elements. Fortunately, those cases are easily dealt with individually.
