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I was just doing some fun reading, and I stumbled upon the following:

A total function $f: X \mapsto Y$ is a binary relation on $X \times Y$ that satisfies the following two properties:

  1. $\forall x \in X$, there is a $y \in Y$ such that $[x, y] \in f$
  2. if $[x, y_1] \in f$ and $[x,y_2] \in f$, then $y_1 = y_2$

That's all nice and dandy, however, the text then goes on to state

A relation on $\mathbb{N} \times \mathbb{N}$ representing greater than fails to satisfy either of the conditions.

I understand why it fails the second condition, but why does it fail the first condition? Isn't it a similar idea to $f: X \mapsto X : x \mapsto x+1$, which would pass the first condition. In other words, isn't there alwasy a number in $\mathbb{N}$ that is greater than the current one?

Cheers!

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  1. For $x = 0$, there is no $y \in \Bbb N$ such that $x>y$.
  2. We have $3 > 1$ and $3 > 2$, but $1 \ne 2$.
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  • $\begingroup$ Wow, realized that I was thinking about less than for rule 1... cheers! $\endgroup$ – agupta231 Jan 21 '18 at 0:25
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  1. The strictly-greater-than relation fails the first condition because there is no natural number $y$ such that $0>y$. (The relation would pass this test if it were defined on all integers $\mathbb{Z}\times\mathbb{Z}$ rather than natural numbers $\mathbb{N}\times \mathbb{N}$.)

    For this reason, there's no function $x \mapsto x-1$ sending each natural number $\mathbb{N}$ to another number in $\mathbb{N}$ that it is greater than.

  2. The strictly-greater-than relation fails the second condition because 1 and 2 are different natural numbers such that $3 > 1$ and $3 > 2$ but $1\neq 2$.

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