In the context of the connected components of a graph, the idea isn't very interesting, because the picture for connected components of a graph is already so clear and easy to understand geometrically.
But one of the principal techniques of mathematics is to find the underlying structure in one context, and then try to make an abstract model of it that might apply in other contexts as well. By doing this, we understand the connections between the two contexts, and we can sometimes solve difficult problems in one context by applying tools imported from some other context.
We don't formulate the idea of a partition or an equivalence relation because we want to study the connected components of graphs. For that it's unnecessary, because graph components are simple.
But that simplicity makes graph theory a good place to start understanding the idea of an equivalence relation, so that when you happen across it in a different context, where it might be more useful, you can recognize it and say “oh, we can model this with an equivalence relation, which means that it partitions the structure into components, and I already know some theorems about how that will work and some techniques I can use.”
And you have a language for talking about these things, which can be applied in many different situations, so that when you say “consider the equivalence classes of (something) under (some relation)” you and the people you are talking to instantly get an idea of what is going on: these classes are disjoint, every object is in exactly one class, and so on, just like the components of a graph.
Examples
What are fractions? They are notations of the form $\frac ab$, where $a$ and $b$ are integers, and $b$ is not zero. What are the rational numbers? Just fractions? No, they are the equivalence classes of fractions under the relation that says $\frac ab \equiv \frac cd$ if $ad=bc$, because for example $\frac 36$ and $\frac8{16}$ are the same rational number. Okay, big deal, we already know all about rational numbers. But having identified the process, we can now apply it to all sorts of things more complicated than the integers. Can we do the same construction for, say, polynomials? (We can!) Are the results useful and interesting? (They are!)
We can relate the complex numbers to polynomials by defining each complex number as a part of a partition induced on the set of polynomials by a certain equivalence relation. By using different equivalence relations we can define different systems analogous to the complex numbers and use them to study properties of polynomials.
Mathematics has a structure called a group, which is a model of a way in which a thing can have symmetry. There is an important “quotient” operation on a group which arises when you consider certain symmetries to be “equivalent”; the quotient is a group that describes the symmetries of the resulting equivalence classes.
This is a very specific example: Consider some geometric object which has an even number of symmetries. Then the object must have at least one symmetry which is an “involution”: this means that if you perform the symmetry exactly twice, the object is back in its original position. Objects without involutions must have an odd number of symmetries! (An example is a table-saw blade with 37 teeth.) I hope this is not obvious! But it is very easy to show if you consider the right equivalence relation on the symmetries.
An important technique in physics is to analyze the symmetries of the universe itself. For example, the laws of conservation of momentum and energy are consequences of certain symmetries of space-time. The abstract structure of spaces with these symmetries is often best understood as a particular quotient group.
This “quotient” idea applies in other situations too. Many kinds of mathematical structures are best understood as quotients, in this sense. For example, in topology we often view a circle as being a quotient of a line segment, under the equivalence relation that says that the two endpoints are equivalent. When we want to deal with a Möbius strip, we often formulate it as a certain quotient of a rectangle, where the equivalence relation makes certain points on the edge of the rectangle equivalent. Many related objects, much weirder than the Möbius strip, can be dealt with similarly.
We don't study the identification between partitions and equivalence relations just because it is cool. It is also useful in understanding other things. Partitions and equivalence relations pop up everywhere. Sometimes the partition is obvious, but the equivalence relation is easier to understand; then it is helpful to reinterpret the partition as an equivalence relation. Sometimes it is useful to go in the other direction instead.
When you defined $a \equiv b \pmod n$ we use the notation $\pmod n$ BECAUSE $a- b\in n\mathbb Z$ is an equivalence relation that means that the relation $a-b\in n\mathbb Z$ "partitions" $\mathbb Z$ into equivalence classes and $\mathbb Z/n\mathbb Z$ is the set of equivalence classes it is partitioned into.
$\mathbb Z/n\mathbb Z = \{[0],[1],......., [n-1]\}$ where $[k]=\{m\in \mathbb Z| m\equiv k \pmod n\} = \{m \in \mathbb Z| m-k \in n\mathbb Z\} = \{....,k-n, k, k+n, k+2n,.....\}$.
But there is nothing magical about the relation $a\equiv b$. EVERY equivalence relation will partition a set, $S$ into equivalence classes $U_\alpha$ where each and every $s\in S$ is in exactly one, and only one, $U_\alpha$ and all the elements in $U_\alpha$ are related to each other. And $S/\sim$ is the set of all these equivalence classes.
So for instance; take any equivalence relation $\sim$ an a set $S$. For example let say $S= \mathbb N$ and $a \sim b\iff $ the highest power of $3$ that divides $a$ is the same highest power of $3$ that divides $b$.
for example $3\sim 6 \sim 12 \sim 51 etc. $ because $3|6, 12, 51$ but $3^2 \not\mid 6,12,51$.
If we let $U_1 = \{3, 6, 12, 15, 21, 24, .....\}=\{k\in \mathbb Z| 3\mid k, 9\not \mid k\}$ then that is one equivalence class. If we let $U_2=\{9,18,36,....\} = \{k\in \mathbb Z| 9|k, 27\not \mid k\}$ that's another equivalence class. And we can let $U_m =\{k\in \mathbb Z| 3^m|m, 3^{m+1}\not \mid k\}$ can be any other.
Also note $\{0\} = U_\infty = \{k\in \mathbb N| 3^j\mid k$ for all $k\in \mathbb N\}$ is the final equivalence class..
Then $\mathbb Z/\sim = \{U_\infty, U_1, U_2, U_3......\}$ is the set of all equivalence classes.
That is entirely equivalent to $\mathbb Z/(\pmod n) = \{[0],[1],.....[n-1]\}$ where $[k] = \{.... k-2n, k-n, k, k+n,...\} = \{k+mn|m\in \mathbb Z\} = \{z\in \mathbb Z| a\equiv z\pmod n\} = \{z\in \mathbb Z| a-z\in n\mathbb Z\}$ are the set of all equivalence classes.
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I do not see any parallel in general to S/∼. The only similarity is that they are equivalent "with respect to ∼."
Thats all the similarity that there is. but that's that's enough similarity for anyone. That is a HUGE similarity.
Best Answer
Solution for (a):
Let $X, Y, Z \in \mathcal P(S)$.
Reflexivity is satisfied, since obviously $X \subseteq X$
Symmetry is not satisfied. We know $S$ has at least two elements, so we can choose two distinct elements $a, b \in S$. Then $\{a\} \subseteq \{a,b\}$ but $\{a,b\} \not \subseteq \{a\}$.
Transitivity is satisfied: Let $X \subseteq Y \subseteq Z$. Now, if $X=\emptyset$, then $X \subseteq Z$. So let $X \neq \emptyset$ and let $x \in X$. Then $x$ is in $Y$ but since $Y \subseteq Z$, we also have $x \in Z$. That shows $X\subseteq Z$.
Can you do the rest now?