Showing equal things are the same object (Tao's Analysis I)

I am reading Tao's Analysis I, and throughout he uses the notion of an object to define even axioms, treating it as an intuitive notion without explaining it. Based on the axioms that he has given for equality, I am trying to reason that if 2 things are equal, they are the same mathematical object, which Wikipedia here says is the case. This is motivated by the comments to this answer here. For context, this is how Tao uses the notion of an object to define the axioms for equality:

• (Reflexive axiom). Given any object $$x$$, we have $$x=x$$.
• (Symmetry axiom). Given any two objects $$x$$ and $$y$$ of the same type, if $$x=y$$, then $$y=x$$.
• (Transitive axiom). Given any three objects $$x, y, z$$ of the same type, if $$x=y$$ and $$y=z$$, then $$x=z$$.
• (Substitution axiom). Given any two objects $$x$$ and $$y$$ of the same type, if $$x=y$$, then $$f(x)=f(y)$$ for all functions or operations $$f$$. Similarly, for any property $$P(x)$$ depending on $$x$$, if $$x=y$$, then $$P(x)$$ and $$P(y)$$ are equivalent statements.

This is my reasoning to show that if 2 things are equal, they are the same object:

if 2 things $$x$$ and $$y$$ are equal, then there doesn't exist a property $$P(x)$$ such that $$P(x)$$ and $$P(y)$$ aren't logically equivalent- otherwise the axiom of substitution would be violated. Therefore $$x$$ and $$y$$ have the same properties and are not discernible in any way. Therefore $$x$$ and $$y$$ are the same object.

I am unsure of this reasoning because it is not clear to me that $$x$$ and $$y$$ are the same object if they have the same properties, as what an 'object' is hasn't been defined at all by Tao. I am also unsure because perhaps the intuitive notion of objects with Wikipedia uses is different than the one Tao uses, meaning the statement I got from Wikipedia about equal things being the same object doesn't hold in the context in which Tao is working: for example, Tao says that "Equality is a relation linking two objects $$x, y$$ of the same type $$T$$", making it sound like 2 equal things don't necessarily have to be the same object, as different objects can be linked through equality.

So does my reasoning to show equal things are the same objects hold in the context in which Tao is working in Analysis I?

• What do you mean by "the same mathematical object" if you don't just mean "equal"? Commented Jun 16 at 0:54
• Frankly, at this stage I recommend you don't worry about this for now and continue reading. The notion of equality in mathematics is actually very subtle and it would take a long time to give a full and satisfying discussion of it (one example of the subtlety: is $2 \in \mathbb{Z}$ "the same" as $2 \in \mathbb{Q}$?). Commented Jun 16 at 1:10
• I am really not sure how to meaningfully address your question. The problem is that, as you say, no formal definition of an "object" has been given, so no formal definition of "the same object" is possible either. Again, I recommend that you don't worry about this for now and continue reading. Commented Jun 16 at 1:28
• Just hope that your intuitive concept is good enough to understand the book for now and if it isn't you can ask a more specific question about where you get stuck later. Commented Jun 16 at 1:34
• @PrincessMia ZF/ZFC might be an example of what you're looking for. In those models, everything is a set, where "set" is defined by a collection of axioms. In particular, the axiom of extensionality defines what it means for two sets to be equal; any definition of equality in these models will eventually reduce to this axiom. Commented Jun 16 at 2:49

Objects are elements of sets.

A relation on a set $$S$$ is any subset of $$S\times S$$.

If $$\sim$$ is a relation on $$S$$ and $$(s_1,s_2)\in \sim$$, then we say that $$s_1$$ is $$\sim-$$related to $$S_2$$ and we write $$s_1\sim s_2$$.

An equivalence relation on a set $$S$$ is a relation $$\sim$$ satisfying three properties, namely:

1. (reflexivity) For all $$x\in S$$, $$x\sim x$$
2. (symmetry) For all $$x,y\in S$$, $$x\sim y$$ if and only if $$y\sim x$$
3. (transitivity) For all $$x, y, z \in S$$, $$x\sim y$$ and $$y \sim z$$ then $$x\sim z$$

so far, this should all sound familiar. If $$\sim$$ is an equivalence relation on a set $$S$$ and $$x\sim y$$ we usually say that $$x$$ is equal to $$y$$, and, if the equivalence relation is understood, then we sometimes write $$x=y$$

Now, the important thing to know here is that every equivalence relation induces a partition of the set $$S$$ into a collection of subsets of $$S$$ called the equivalence classes of $$S$$ with respect to $$\sim$$.

A partition $$P$$ of a set $$S$$ is a collection of subsets of $$S$$ (that is $$P\subset 2^{S})$$ satisfying three conditions, namely:

1. For each $$a\in P$$ there exists an $$x\in S$$ such that $$x\in a$$ (i.e a is nonempty*)
2. $$\underset{a\in P}\bigcup a=P$$ (i.e. $$P$$ covers $$S$$)
3. For distinct sets $$a,b\in P$$ $$a\cap b=\emptyset$$ (i.e. the elements of P are pairwise disjoint)

Now, if $$\sim$$ is an equivalence relation on a set $$S$$ and $$x\in S$$ we define the equivalence class of $$x$$ with respect to $$\sim$$ to be the set $$[x]:=\{y\in S| y\sim x\}$$ and we denote the set $$\{[x]|x\in S\}$$ by $$S/\sim$$

Now, it is a very easy exercise to show that every equivalence relation $$\sim$$ on a set $$S$$ induces a partition of $$S$$, namely $$S/\sim$$ and conversely, every partition $$P$$ of $$S$$ induces an equivalence relation on $$S$$, namely the relation defined by $$x\sim y$$ if and only if there exists an $$a\in P$$ such that $$x\in a$$ and $$y \in a$$

In conclusion, we say that two "objects" $$x$$ and $$y$$ in a set $$S$$ are "equal" when there is an implied equivalence relation $$\sim$$ on $$S$$ and $$x\sim y$$, or equivalently, when $$[x]=[y]$$. This does not necessarily mean that $$x$$ and $$y$$ are the same objects. That depends on how you define "sameness".

For example, we can define the rational numbers to be a set of equivalence classes of ordered pairs of integers where $$(a,b)$$ is identified with $$\frac{a}{b}$$ and the equivalence relation is $$(a,b)\sim (c,d)$$ if and only if $$ad=bc$$, with addition and multiplication defined by

$$(a,b)+(c,d)=(ad+bc, bd)$$ and

$$(a,b)\times(c,d)=(ac, bd)$$

Then

$$\mathbb{Q}:=\mathbb{Z}\times\mathbb{Z}\backslash \{0\}/\sim$$

In this case, $$[(1,2)]=[(2,4)]$$, and this is just the statement that $$\frac{1}{2}=\frac{2}{4}$$. Its not that $$(1,2)$$ and $$(2,4)$$ are the same object, it just means that they are in the same equivalence class. Therefore we say that $$\frac{1}{2}$$ and $$\frac{2}{4}$$ are equal because they represent the same equivalence class. However when we think of these two ordered pairs as rational numbers we usually do think of them as "the same" number.

I hope this helps :)

• Thank you! I noticed that equivalence relations do not need to satisfy the axiom of substitution, whereas equality needs to satisfy the axiom of substitution. Doesn't this mean that equality and being the same though an equivalence relation are different? Commented Jun 16 at 4:09
• @PrincessMia When I can, I'll update my answer to address this, in the mean time check out this Commented Jun 16 at 4:14