# 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!

• are both triangles is not an equivalence relation, as a circle does not relate to itself, contradicting reflexivity. Not a square is a unary relation, so it trivially is not an equivalence relation. None of the examples you've provided are equivalence relations. Make sure you actually check the properties, not "I think this is reflexive, symmetric, transitive" Commented Dec 21, 2021 at 15:55
• It's also worth considering the set over which the relation is defined. Certain relations will not partition the set you are considering, simply because they do not touch certain elements. Commented Dec 21, 2021 at 16:06
• Two equivalence classes from two different equivalence relations can intersect. An equivalence relation is equivalent to a partition of the set. If you don't understand these concepts, you are not ready for topology. Commented Dec 21, 2021 at 16:36
• To get an equivalence relation, as noted above the relation "are both triangles" won't do because it's not reflexive. The relation you want, which is reflexive, is "are both the same shape." Commented Dec 21, 2021 at 16:41
• My main errors in understanding were (1) thinking that multiple equivalence relations could be applied to partition a set. Only one at a time can be used. I think I got confused because you can have multiple equivalence classes, but one equivalence relation is used to form those classes, and (2) the equivalence relation has to be reflexive, symmetric and transitive AND has to make sense for all the elements in that set as William and Robert Shore explain. [Comment for future self-learners :)]. Thank you all! Without a professor or classmate, this site is essential for me!
– Dave
Commented Dec 23, 2021 at 12:20

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.

• The other comments above are right; I was a bit too quick with interpreting your question. You should change the definition of $\sim_1$ and $\sim_2$ and make sure that they define equivalence relations. I would define $\sim_1$ by: $a\sim_1 b$ if and only if $a$ and $b$ are both squares or both not squares. Similarly, if you define $\sim_2$ by $a\sim_2b$ if and only if $a$ and $b$ are both triangles or if $a$ and $b$ are both not triangles, then my answer works. Commented Dec 21, 2021 at 16:18
• Thank you very much SilverBlueZ and sorry for the late reply. You cleared it up for me immensely. I was confused by many things but the two key things were, (1) I did not realize you need to consider just one relation when partitioning, and (2) the relation needs to make sense for all the elements. As Robert Shore pointed out above, "are both triangles" doesn't make sense for all elements so we don't have reflexivity. In retrospect, this is all very easy. Unfortunately with self-study, it's easy to go down the wrong path, when a 3 minute conversation would clear it up, so Thank you!
– Dave
Commented Dec 23, 2021 at 12:12

"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".