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I made a typo, so my question is repeating, but it is edited.

Can someone help me to solve this problem?

Let $(M,d)$ be a metric space and let function $d$ be only $0, 1,3$. We say that element $x \in M $ is equivalent with element $y \in M $ if $d(x,y )\leq 1 $. Let $H$ be a set of all ordered pairs $\left ( x,y \right )\in M^{2} $, which are equivalent. Prove that $H$ is equivalence relation on set $M$.

So I need to verify that H is reflexive, symetric and transitive. But how?

  • What happened with, say, the reflexive property, when you wrote out the definition of reflexive and then used that definition to write out what you may assume and what you must prove? In general, can you tell us what you know and what have you tried? It's hard to know on what level to help you if we don't even know whether you know the basic definitions and how to apply them. – Lee Mosher Feb 16 '20 at 17:19

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For $x,y\in M$, let $x\sim y$ iff $d(x,y)\le1$.

Let $x\in M$. Then $d(x,x)=0$. Hence $x\sim x$. So $\sim$ is reflexive.

Let $x,y\in M$. Note that $d(x,y)=d(y,x)$. So $d(x,y)\le1$ iff $d(y,x)\le1$. Hence if $x\sim y$, then $y\sim x$. So $\sim$ is symmetric.

Let $x,y,z\in M$ with $x\sim y$ and $y\sim z$. So $d(x,y)\le1$ and $d(y,z)\le1$. So by the triangle inequality we have that:

$$d(x,z)\le d(x,y)+d(y,z)\le2$$

But since $d(x,z)\le2$, we must also have that $d(x,z)\le1$, since $d(x,z)$ can only be either $0$, $1$, or $3$. Hence $x\sim z$, and $\sim$ is transitive.

user729424
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