Equivalence Relations from $\mathbb Z \to\mathbb Z$ The question is, "Which of these relations on the set of all functions from $\mathbb Z$ to $\mathbb Z$ are equivalence relations."
The first relation to consider is $\{(f,g)|f(1)=g(1)\}$,it seems easy ,but I don't know how to handle
The second relation to consider is, $\{(f,g)|f(0)=g(0)∨f(1)=g(1)\}$ For this one, I am not very certain where to begin.
The third one, $\{(f,g)|f(x)−g(x)=1,\forall x∈\mathbb Z\}$ I can see how this isn't reflexive, because f(x)−f(x)=0 is always true for any function. I can also see how it isn't symmetric, because although $f(x)−g(x)=1$ could be true, it's counter-part, $g(x)−f(x)=−1$, won't be true. Despite me being able to see those facts, I can not see how it isn't transitive. How would I show that?
The fourth one, $\{(f,g)|$for some $C ∈\mathbb Z,\forall x∈\mathbb Z,f(x)−g(x)=C\}$ I had the idea that it wasn't an equivalence relation based on the fact that if we let $f(x)=x$, and $g(x)=x−1$, and say x=1, then $f(1)−g(1)=C\implies C=1$, but $g(1)−f(1)=C\implies C=−1$. The C values aren't the same, implying the relation wouldn't be symmetric, right? Also, I am not sure how to prove or disprove that the relation is transitive.
The last one is similar to the first one: $\{(f,g)|f(0)=g(1)∧f(1)=g(0)\}$; and like the first one, I am not sure where to begin.
Sorry that this post is rather long. But thank you for reading! and I hope that you can help me.
 A: *

*For the first one, $(f,f)$ always belongs to the relation because $f(1)=f(1)$. Hence it is reflexive.
 It is symmetric too, as $f(1)=g(1)\implies g(1)=f(1)$. And of course, it is transitive, as $$f(1)=g(1)\land g(1)=h(1)\implies f(1)=h(1),$$ thus causing $(f,h)$ to belong to the relation whenever $(f,g)$ and $(g,h)$ both do. This has to do directly with Equality being an equivalence relation.

*This is almost identical to the first one.

3,4. You might get some clarity by observing that the fourth one is a direct generalisation of the third one, with a parameter C replacing the constant $1$. For transitivity, if $(f,g)$ and $(g,h)$ belong to the set, then $f(x)-g(x)=g(x)-h(x)=1$ and adding them gives $f(x)-h(x)=2$, which means $(f,h)$ definitely cannot belong to the relation. Thus, non-transitive.


*This one is not reflexive, as all ordered pairs $(f,f)$, where $f:\mathbb Z\to \mathbb Z$ is any function, do not belong to the relation set. It is symmetric though, which should be easy to see. For transitivity, if $(f,g)$ and $(g,h)$ belong to the set, then we have $$f(0)=g(1), g(0)=f(1)$$$$g(0)=h(1), h(0)=g(1).$$ which means that $$f(0)=h(0), f(1)=h(1)$$ which means that $(f,h)$ needn’t necessarily be in the set. Thus non-transitive.

A: Keep in mind that the equivalence property is actually $3$ different properties.

Definition 1: Let $R$ be a relation on a set $A$. We say that $R$ is Reflexive  iff $\forall x\in A$,
$$xRx$$


Definition 2: Let $R$ be a relation on a set $A$. We say that $R$ is Symmetric  iff $\forall x\in A, \forall y\in A$,
$$xRy \Rightarrow yRx$$


Definition 3: Let $R$ be a relation on a set $A$. We say that $R$ is Transitive  iff $\forall x\in A, \forall y\in A, \forall z\in A$,
$$xRy \text{ and }yRz\Rightarrow xRz$$

So to prove a relation is an equivalence relation, you need to prove all three properties. Let us denote $\mathcal{C}$ to be the set of all auto functions on $Z$
Let us start with the first relation on $\mathbb{Z}$ we say that
$$fRg\Leftrightarrow f(1)=g(1)$$
Reflexive: $\forall f\in \mathcal{C}$,
$$f(1)=f(1)$$
$$\Rightarrow fRf$$
$$\Rightarrow R \text{ is reflexive }$$
Symmetric: $\forall f\in\mathcal{C},\forall g\in \mathcal{C}$,
$$fRg$$
$$\Rightarrow f(1)=g(1)$$
$$\Rightarrow g(1)=f(1)$$
$$\Rightarrow gRf$$
$$\Rightarrow R \text{ is symmetric}$$
Transitive: $\forall f\in\mathcal{C},\forall g\in \mathcal{C}, \forall h\in \mathcal{C}$,
$$fRg\;\wedge \; gRh$$
$$\Rightarrow f(1)=g(1)\; \wedge \; g(1)=h91)$$
$$\Rightarrow f(1)=h(1)$$
$$\Rightarrow fRh$$
$$\Rightarrow R \text{ is Transitive}$$
hence why we concude that $R$ is an equivalence relation.
As for the second relation
$$fTg\Leftrightarrow f(0)=g(0)\;\vee\; f(1)=g(1)$$
this relation is not transitive.
Let $f(x)=x, g(x)=2x, \text{ and } h(x)=x+1$
$$f(0)=g(0)=0, g(1)=h(1)=2, f(1)=1, \text{ and }h(0)=1$$
$$\Rightarrow f(0)\ne h(0)\text{ and }f(1)\ne h(1)$$
$$\Rightarrow fTg\;\wedge \; gTh \text{ but } f\not T h$$
which means that $T$ is not transitive and, hence not an equivalence relation.
As for the third and forth relation, these will only be true if $C=0$ which means they are the same function. For $C\ne 0$
$$fKg\Leftrightarrow \forall x\in \mathbb{Z}, f(x)-g(x)=C$$
This relation is not reflexive ( nor transitive but reflexive is easier to disprove)
Let $f(x)=x$, $\forall x\in \mathbb{Z}$
$$f(x)-f(x)=0\ne C$$
$$\Rightarrow f\not K f$$
which means that $K$ is not reflexive and, hence not an equivalence relation.
As for the last relation
$$fPg\Leftrightarrow f(0)=g(1)\;\wedge\; f(1)=g(0)$$
Again this one is not reflexive. Let $f(x)=x$ then
$$f(0)=0\;\wedge\; f(1)=1$$
$$\Rightarrow f(0)\ne f(1)$$
$$\Rightarrow f\not P f$$
which means that $P$ is not reflexive and, hence not an equivalence relation.
Hope this helped.
