$Q1$
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 $(0,0)$.
In the argument for symmetry, you've shown that if $f = g$ then $f \sim g$ and $g \sim f$. But this does not guarantee symmetry, which requires that if $f \sim g$ then $g \sim f$. To do this, $f \sim g$ shows that there is a constant $k$ such that $f(x) = g(x)$ for all $x \geq k$, and you need to show that there is a constant $l$ such that $g(x) = f(x)$ for all $x \geq l$.
For transitivity, you must show that if $f \sim g$ and $g \sim h$ then $f \sim h$. In your case, this means there are constants $k$ such that $f(x) = g(x)$ for $x \geq k$ and $l$ such that $g(x) = h(x)$ for $x \geq l$, and you need to show that there is a constant $m$ such that $f(x) = h(x)$ for all $x \geq m$. How might you produce such an $m$ in terms of the other data you have available?
Best Answer
For transitivity to hold we need $(xy\ge 1\land yz\ge 1)\implies (xz\ge 1)$ for all $x,y,z.$ But it's false when $x=z=1/2$ and $y=2.$