I am trying to determine if the following sets are equivalence relations:
Set: $B=\mathbb{R}$
Relation: $x \sim y$ if $|x-y|\leq 1$
Reflexive property: For $x \sim x$, we have $|x-x|\leq 1=|0|=0\leq1$ so this is true.
Symmetry property: We need to ensure that if $x \sim y$ then $y \sim x$. This is also true since we have $|y-x|=|-(x-y)|=|x-y|$ and we know that $|x-y|\leq1$ so symmetry is satisfied.
Transitive property: I always struggle with this part. I think what I need to show is that if $|x-y|\leq1$ and $|y-z|\leq1$, then $|x-z|\leq1$. If my assumption is right, how would I go about doing this? My idea was to write $|x-z|\leq1$ as $|(x-y)+(y-z)||\leq1$. Since $(x-y)$ and $(y-z)$ could both potentially equal something that adds to a value larger than 1, transitivity is NOT satisfied so this is not an equivalence relation. However, I am not sure if this is the right approach...
Set: $B=M_n(\mathbb{R})$, the set of $n \times n$ matrices with real entries, where $n$ is some fixed positive integer.
Relation: $X \sim Y$ if $Tr(X-Y) \in\mathbb{Z}$
Reflexive property: For $x \sim x$ we have $Tr(X-X)=0\in \mathbb{Z}$ so the reflexive property is satisfied.
Symmetry property: We need to ensure that if $x \sim y$ then $y \sim x$. We have $Tr(Y-X)=-Tr(X-Y)$ which is still in $\mathbb{Z}$ so symmetry is satisfied
Transitivity: Again, this is what I struggle with. I believe that here, we need to check that if $Tr(X-Y)\in \mathbb{Z}$ and $Tr(Y-Z)\in \mathbb{Z}$, then we need to show that $Tr(X-Z)\in \mathbb{Z}$.
My idea was to write $Tr(X-Z)$ as $Tr(X)-Tr(Z)$. However, I am not sure if this implies that $Tr(X)\in \mathbb{Z}$ and $Tr(Z)\in \mathbb{Z}$ which implies transitivity.
Any guidance on this would be very much appreciated! Also, is my work on the other properties right?
I'll answer this in two parts, as you have asked.
$B = \mathbb R$, and $\sim$ is such that $x\sim y$ if $|x-y| \le 1$. The relation $\sim$ is reflexive and symmetric as you've shown. However, it is not transitive.
To see this, consider $0,1,2\in\mathbb R$. $0\sim 1$ since $|0-1| = 1 \le 1$. Also, $1\sim 2$ since $|1-2| = 1 \le 1$. Sadly, $0 \not\sim 2$, since $|0-2| = 2 > 1$.
$B = M_n(\mathbb R)$, and $\sim$ is such that $X\sim Y$ if $\text{trace}(X-Y) \in \mathbb Z$. Once again, you're right about the reflexivity and symmetry parts. I shall show transitivity. Suppose $X\sim Y$ and $Y\sim Z$. Thus, $\text{trace}(X-Y) \in\mathbb Z$ and $\text{trace}(Y-Z) \in \mathbb Z$. Since the sum of two integers is an integer, we have $\text{trace}(X-Y) + \text{trace}(Y-Z) \in \mathbb Z$. Note that in general, for any two square matrices $A$ and $B$ of the same order, $\text{trace}(A+B) = \text{trace}(A) + \text{trace}(B)$. So, $$\text{trace}(X-Y) + \text{trace}(Y-Z) = \text{trace}(X) - \text{trace}(Y) + \text{trace}(Y) - \text{trace}(Z)\\=\text{trace}(X) - \text{trace}(Z) = \text{trace}(X-Z)$$ Hence, $\text{trace}(X-Z)\in\mathbb Z$ and $X\sim Z$ as required. Thus, $\sim$ is an equivalence relation (reflexive, symmetric and transitive).