I need help understanding Equivalence Relations and Equivalence Classes to proceed in Topology

I need help understanding equivalence classes and relations. I'm self-studying topology (General Topology by Willard, Introduction to Topology by Mendelson). I have read the explanations in the book, searched Wikipedia ( https://en.wikipedia.org/wiki/Equivalence_class) and this site but am still confused.

Specifically I'm confused by the statement on page 16 of Mendelson: "Two equivalence classes are either disjoint or identical.". A similar statement appears in Willard and elsewhere so I suspect the statement is correct :). Following from this is the statement: "the equivalence classes form a partition of S." (Wikipedia) It's possible I'm confused by this bit because I don't understand the preceding discussion as well as I think. An example will make clear my confusion (I hope :)).

Consider a set S consisting of various triangles, squares and circles, could I not define a relation "are both triangles"? And another: "are both circles" and "are both squares". Each of these relations is reflexive, symmetric and transitive. And I can see that these classes are disjoint AND form a partition of S.

BUT I could also consider a relation "Not a square". Two members have the relation if they are both not squares. Again, I think this is reflexive, symmetric and transitive. But, it absolutely is not the case that "Not a Square" and "Are Both circles" are "disjoint or identical". And I don't see how we can say S is partitioned with this additional equivalence class.

I feel I'm missing something basic. Must we consider all possible equivalence classes? Or are there some other rules I'm missing/misunderstanding? Perhaps a simple example will clear up my misunderstanding.

Thanks!

There is something fundamental going wrong in your reasoning. You are defining two different relations on $S$:

1. $a\sim_1 b$ if and only if $a$ and $b$ are both not squares.
2. $a\sim_2 b$ if and only if $a$ and $b$ are both circles.

However, these relations are not equivalence relations yet; there are elements in $S$ that do not relate to itself, so it is not reflexive. There are many ways to modify $\sim_1$ and $\sim_2$ such that it does define an equivalence relation. For instance:

1. $a\sim_1 b$ if and only if $a$ and $b$ are both squares or are both not squares.
2. $a\sim_2 b$ if and only if $a$ and $b$ are both circles or $a=b$.

Now, both relations are transitive, symmetric and reflexive and therefore equivalence relations. However, each of the relations defines its own partition of $S$.

For instance, $\sim_1$ partitions $S$ into two equivalence classes: the set $A$ that consists of all the elements of $S$ that are not squares and the set $B$ that consists of all elements of $S$ that are squares. Note that $A$ and $B$ are disjoint sets! (This is what is meant).

Similarly, $\sim_2$ partitions $S$ into many classes: the class that contains all circles and each element in $S$ that is not a circle defines its own equivalence class consisting of just itself. These classes are obviously disjoint as well.

What you did instead is compare equivalence classes defined by $\sim_1$ with classes defined by $\sim_2$.

I hope this helps.

"Two equivalence classes are either disjoint or identical."

This can be proved by application of the definition of an equivalence relation: it is reflexive, symmetric and transitive.

Let $R$ be the equivalence relation.

Let $S$ and $T$ be equivalence classes under $R$.

Suppose $S$ and $T$ are not disjoint.

Then $\exists x \in S \cap T$.

That is, $x \in S$ and $x \in T$.

Suppose $y \in S$.

Then $(x, y) \in R$ by definition of equivalence class.

But $x \in T$ and $T$ is an equivalence class of $x$.

So by definition of equivalence class, $y \in T$.

So $y \in S \implies y \in T$ so $S \subseteq T$.

Similarly by making the assumption $z \in T$ it follows that as $x \in T$ we have $(z, x) \in R$.

But $x \in S$ and $S$ is an equivalence class of $x$.

So by definition of equivalence class, $z \in S$.

So $z \in T \implies z \in S$ and so $T \subseteq S$.

We have that $S \subseteq T$ and $T \subseteq S$ so $S = T$.

We have shown that if $S$ and $T$ are not disjoint, then $S = T$.

The rest of your post suggests that you may need to think deeply about the nature of set partitioning.

Talking about an equivalence class is just a way to partition a set into (disjoint) subsets. Being equivalent means belonging to the same subset.

The relation you define:

Both not squares is not an equivalence relation. If $A$ and $B$ are not both squares and $B$ and $C$ are not both squares, can you conclude that $A$ and $C$ are not both squares???

Two elements $a$ and $b$ are related ($a \sim b$) if there is a set in the partition to which both of them belong. This already means that $a \sim b \Leftrightarrow b \sim a$.

A partition $\mathcal{P}$ satisfies:

1. Cover the set: $$X = \bigcup_{P \in \mathcal{P}} P.$$ That is, $a \sim a$.
2. Be formed by disjoint sets. That is, $$a \sim b, b \sim c \Rightarrow a \sim c.$$

In your "both not squares" relation, if you think of $$\mathcal{P} = \{C_a |\, a \in X\},$$ where $$C_a = \{b \in X |\, a \sim b\},$$ this does not partition the set because the sets are not disjoint. And this happens exactly because there are squares $a$ and $c$ and a triangle $b$, and $$ \begin{align*} b &\in C_a \cap C_c, &(b \sim a\text{ and }b \sim c) \\ C_a &\neq C_c. &(a \not \in C_c\text{, because } a \not \sim c) \end{align*} $$ Transitivity does not hold exactly when the sets are not disjoint.

And as @SilverBlueZ points out, by the way, $a \not \sim a$. That is, $a \not \in C_a$.

This means that although you might even be partitioning the set, the relation cannot be formulated in terms of "both belong to the same partition subset".