Why does $a≤b{\implies}a=b∧a<b$?

Question

Let $≤$ be a relation on set $A$ such that $≤$ is reflexive, transitive and antisymmetric.

Let $<$ be a relation on set $A$ such that $<$ is asymmetric, antireflexive and transitive.

Let $a$ and $b$ be two different elements of $A$. Why is it the case that if $a≤b$, then either $a=b$ or $a<b$? Why can't it be neither?

Answer 1

The $\vee$ symbol means OR and the $\wedge$ symbol means AND.

Remember that $\wedge$ looks like $\cap$ which means "In this set AND in that set"

On the other hand $\vee$ looks like $\cup$ which means "In this set OR in that set".

Or you could remember that $\wedge$ looks like the A from "And".

You have written $(a≤b){\implies}(a=b)∧(a<b)$

That means $(a \le b) \implies (a=b$ AND $a < b )$.

The second part of that cannot be true because if $a$ is equal to $b$ then $a$ cannot be less than $b$.

However if we wrote $(a \le b) \implies (a=b$ OR $a < b )$ then that would be true.

After all, what does $a \le b$ mean if not "$a$ is less than or equal to $b$"?

Answer 2

Let a and b be integers. Let a $\le$ b be defined as a | b thats reflexive, transitive and antisymmetric.

Let a < b be such that a is strictly larger than b. That's not reflexive , it's assymetric , and it's transitive.

Now $2 |4$ but $2 \ne 4$ and 2 is not strictly larger than 4 either so your statement is simply not true.

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Unless you actually meant $\le$ to mean the standard "less than OR equal" and < to mean "less than"...

In which case "less than OR equal" means "less than" or "equal" because that's the definition!!!!

Answer 3

Now that I understand you confusion better.

The linked page defines a "strict order" relation. The most well known strict order is "less than" where a is "less than" b iff b-a is positive. But a strict order can be any relationship that is irreflexive, asymmetric, and transitive. Other possible strict orders could be "greater than"; a is "greater than" b iff a -b is negative. Another is "is a proper integer divisor of".

Abstractly, we can use any such relationship to order a set. And abstractly we can use the symbol "<" the indicate this strict order. If we use this symbol we must define what we mean.

Now a strict order can have an additional charactoristics of being "total". That means for any two non-equal elements a and b, a and b must be related either a < b or b < a. "less than" is a total relation. "Is a proper factor of" is not. Either a is less than b or b is less than a (or the are equal). "Less than" us a total order. "Is a factor of" is not. It's possible for two different integers to have neither be a factor of the other.

Then there is the definition of a "partial order". A partial order is a relationship that is reflexive. It is antisymmetric (not assymetric). And it is transitive. "Less than or equal to" and "divides (and could be equal)" are two examples of partial orders.

Now an arbitrary strict order such as "less than" and an arbitrary partial order such as "divides" don't need to have anything in commom.

BUT this is your contention. You can take any total order "<" and define a partial order "$\le $" by defining a $\le $ b iff either a < b or a = b. Example: if I defined "<" to be the strict order "is a proper factor of", I can define the partial order "$\le $" as "is a proper factor or is equal to".

And I can take a partial order "$\le $" and define a strict order by defining a < b iff a $\le$ b and a $\ne $ b.

But if I do define theses partial and strict orders, I define them in terms of each other. I can not say if R is a strict order and S is a totally unrelated partial order than if a S b but a $\ne$ then a R b. That'd be simply nutty.

But if "$\le $" was defined to mean "either < or =", then I could conclude if a $\le $ b and a $\ne $ b then a < b. Because that was the definition.