The order in (1) is known as the lexicographic order on $\Bbb N\times\Bbb N$ and is actually a total order. To show that it’s antisymmetric, suppose that $\langle m_1,n_1\rangle\le\langle m_2,n_2\rangle$ and $\langle m_2,n_2\rangle\le\langle m_1,n_1\rangle$.
- Since $\langle m_1,n_1\rangle\le\langle m_2,n_2\rangle$, we know that either $m_1<m_2$, or $m_1=m_2$ and $n_1\le n_2$.
- Since $\langle m_2,n_2\rangle\le\langle m_1,n_1\rangle$, we know that either $m_2<m_1$, or $m_2=m_1$ and $n_2\le n_1$.
We cannot have $m_1<m_2$: that would imply that $m_2\not<m_1$ and $m_2\ne m_1$, contradicting (2). Thus, $m_1=m_2$, and $n_1\le n_2$. And from (2) we can then see that $n_2\le n_1$ as well, since $m_2=m_1$. Finally, the usual order on $\Bbb N$ is antisymmetric, so $n_1\le n_2$ and $n_2\le n_1$ imply that $n_1=n_2$. We’ve now shown that $m_1=m_2$ and $n_1=n_2$, i.e., that $\langle m_1,n_1\rangle=\langle m_2,n_2\rangle$, as desired.
In the second problem you’re given a transitive relation $<$ on some set $X$, and you’re further told that $<$ has the trichotomy property: for any $x,y\in X$, exactly one of the statements $x<y$, $x=y$, and $y<x$ is true. Now you’re to define a new relation, $\le$, on $X$ by setting $x\le y$ if and only if either $x<y$ or $x=y$. Finally, you’re to prove that this new relation $\le$ is a total order.
To do this you must show that $\le$ is reflexive, antisymmetric, transitive, and total. Reflexivity and transitivity are pretty straightforward, and I’ll leave them to you. For antisymmetry, suppose that $x\le y$ and $y\le x$. Since $x\le y$, we know that either $x<y$ or $x=y$. Suppose that $x\ne y$. Then $x<y$, and by the trichotomy property we know that it is not the case that $y<x$. But then we have $y\not<x$ and $y\ne x$, so be definition it’s not the case that $y\le x$. This contradiction shows that we cannot have $x\ne y$ and hence that $x=y$, as desired. Finally, to show that $\le$ is total, use trichotomy again: for any $x,y\in X$, either $x<y$, in which case $x\le y$; or $x=y$, in which case $x\le y$; or $y<x$, in which case $y\le x$.
This exercise shows how you can turn any so-called strict linear (or total) order $<$ on a set $X$ into an ordinary total order $\le$ in such a way that the $\le$ symbol has the obvious, expected meaning.
This is just the opposite of the previous problem: starting with a total order $\le$ on $X$, turn it into a strict linear order by ‘throwing away the equals part of it’. That is, given the total order $\le$ on $X$, define a new relation $<$ on $X$ by $x<y$ if and only if $x\le y$ and $x\ne y$. Now you’re to show that this new relation is transitive and has the trichotomy property: for any $x,y\in X$, exactly one of $x<y$, $x=y$, and $y<x$ holds. Proving transitivity is very straightforward, and I’ll leave it to you. Trichotomy isn’t much harder, but you do have to check two things: you must show that for any $x,y\in X$ at least one of $x<y$, $x=y$, and $y<x$ holds, and you must also show that it’s never the case that more than one holds. To see, for example, that it’s not possible to have both $x<y$ and $y<x$, note that $x<y$ implies that $x\le y$, and $y<x$ implies that $y\le x$. The relation $\le$ is antisymmetric, so $x=y$. But then by definition $x\not<y$, and we have a contradiction. I’ll leave the rest to you, but feel free to ask for help if you get stuck.
$Q1$
Can a relation be both partial order and equivalence?
Yes, for example, the equality relation.
Is $R_1$ Transitive?
No. It has $(1,0)$ and $(0,7)$ but not $(1,7)$. As this example show, if you add an ordered pair to a transitive relation it can become non-transitive.
A relation on set $A$ that is both reflexive and transitive but neither an equivalence relation nor a partial order (meaning it is neither symmetric nor antisymmetric) is:
$$R_3 = \left\{(0,0),\, (7,7),\, (1,1),\, (0,7),\, (7,1),\, (0,1),\, (1,7) \right\}$$
Reflexive? Yes, because it has $(0,0),\, (7,7),\, (1,1)$.
Transitive? Yes. We go through the relevant cases:
$$(0,7) \mbox{ and } (7,1) \Rightarrow (0,1) \qquad\checkmark$$
$$(7,1) \mbox{ and } (1,7) \Rightarrow (7,7) \qquad\checkmark$$
$$(0,1) \mbox{ and } (1,7) \Rightarrow (0,7) \qquad\checkmark$$
$$(1,7) \mbox{ and } (7,1) \Rightarrow (1,1) \qquad\checkmark$$
Symmetric? No, because we have $(0,1)$ but not $(1,0)$
Antisymmetric? No, because we have $(1,7)$ and $(7,1)$.
$Q2$
Your relation, $R_2$, is correct but your explanations for symmetric and antisymmetric are the wrong way around.
$R_2$ is not antisymmetric because there is as two-way street between distinct vertices, namely, $0$ and $7$.
$R_2$ is symmetric because there is no one-way street between distinct vertices.
Also, $R_2$ is not transitive because it has $(0,7)$ and $(7,0)$ but not $(0,0)$.
Best Answer
As you observed, any of these "less than" relations is NOT a partial ordering — exactly because it's not reflexive. They are antisymmetric and transitive, though — which, of course, shouldn't be taken for granted, but should be verified for each example. But then we can define another "less than or equal to" relation, which will be a partial ordering. In other words, each of the examples above talks about two different (but closely related, no pun intended) relations — "less than" and "less than or equal to".
I can agree that this text is a bit sloppy at times not to distinguish them clearly enough. For example, in the end of your first screenshot, the phrase
appears to be talking about the "less than" relation, for which it isn't true — of course, the question is actually about the "less than or equal to" version. (Maybe it is made clear in the exercise?)
In other words, you can think of this as a two-step definition.
P.S. And no, you can't prove reflexive from antisymmetric, as these are two different properties, and neither implies the other.