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After reading the following post (Why does one have to check if axioms are true?), I wanted a reality check to make sure that I am thinking about equivalence relations correctly.

In Tao's Analysis I, the following statements are made:

Two ordered pairs $(x,y)$ and $(x',y')$ are considered equal if and only if both their components match, i.e.$(x,y) = (x',y') \iff x=x' \text{ and } y=y'$. This obeys the usual axioms of equality (Exercise 3.5.3).

Exercise 3.5.3 reads as follows:

"Show that the definitions of equality for order pair [...] obey the reflexivity, symmetry, and transtivity axioms".

I am looking for confirmation that the following thinking is the correct sequential process one goes through when determining equality between two objects.

Imagine that I just discovered/invented the concept of an ordered pair and don't know anything about them. I subsequently state the following:

$(x,y) R_{op} (x',y') := (x=x' \land y=y')$, where the parentheses on the right are strictly being used to "separate" the symbology from the "defined as" symbol, :=.

i.e. I am first defining what the relation $R_{op}$ means. (where "op" is short for ordered pair).

It dawns on me that this relationship that I just defined could, possibly, satisfy the 3 axioms of equality. I go through and confirm these three statements:

Reflexivity: $(x,x) R_{op} (x,x)$

  1. $(x=x \land y=y)$

Symmetry: if $(x,y) R_{op} (x',y')$, then $ (x',y') R_{op} (x,y)$

  1. if $(x=x' \land y=y')$, then $(x'=x \land y'=y)$

Transitivity: if $(a,b)=(c,d) \text{ and } (c,d)=(e,f)$ then $(a,b) = (e,f)$

  1. if $(a=c \land b=d) \land (c=e \land d=f)$, then $(a=e \land b=f)$

I then say, "Oh, man. This relationship satisfies all three axioms of equality"

Consequently, I can "promote" $R_{op}$ to an equivalence relation, which allows me to use the symbol "$=$" instead of "$R_{op}$". Now, and only now, can I write what Tao writes, which is:

$(x,y) = (x',y') \iff x=x' \text{ and } y=y'$

And, really, I think the more pedantic way of formulating this is:

$(x,y) = (x',y') \iff (x,y) R_{op} (x',y')$, where $(x,y) R_{op} (x',y') := (x=x' \land y=y')$

Is this procedurally correct? Thank you!

S.C.
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    Seems okay to me. – Somos Mar 05 '20 at 23:17
  • Too complex. . . – William Elliot Mar 06 '20 at 01:51
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    @WilliamElliot If by "complex" you mean "too many steps to outline a concept", then sure, I certainly agree. But for someone that is quite a math novice, if this is how equivalence relations were initially explained to me, I would have comprehended the concept immediately, rather than struggling with it for several months. I feel that, many times, the "simple / non-complex" explanations that mathematicians provide often omit too much detail...and therefore omit too much underlying intuition for why things are the way that they are. i.e. what you find "pretty", I likely find inaccessible. – S.C. Mar 06 '20 at 02:07
  • See my answer with that notational baggage removed. @S.Cramer – William Elliot Mar 06 '20 at 09:15
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    "Now, and only now, can I write what Tao writes...". IMO you have not understood the Exercise. You already know that $=$ between "basic" objects: $x,y$ has the three properties. You have to check that the $=$ defined in the way it is defined between the new objects $(x,y)$ has the said properties. – Mauro ALLEGRANZA Mar 06 '20 at 12:15

1 Answers1

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Reflectivity
(x,y) = (x,y) iff x = x and y = y

Symmetry (x,y) = (a,b) iff x = y and a = b
iff a = b and x = y iff (a,b) = (x,y)

Transitivity. Assume (a,b) = (u,v) and (u,v) = (x,y).
Thus a = u and b = v and u = x and v = y.
Hence a = x and b = y; (a,b) = (x,y).

William Elliot
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    Thank you for the answer, but the question was not about the exercise. I only included the details of the exercise because Tao referenced the exercise. The exercise is straightforward...as you just depicted. – S.C. Mar 06 '20 at 14:09
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    @S.Cramer. Yes your work was correct. What I've said before is that your proof is uselessly complex and as I've say in a reply, I would show you a unencumbered proof which answered your question about procedural correctness. No your proof is not procedurally even though logically correct because the procedure you used is inefficient. This answer is the promised procedural correct method that you were asking about. – William Elliot Mar 06 '20 at 19:16