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I saw the following example of an equivalence relation in a topology textbook:

Let $X$ be the unit square. Define $\sim$ as follows:

$(0,y)\sim(1,y)$ for all $y\in [0,1]$, and for any point $(x,y)\in X, (x,y)\sim(y,x).$

How would we show that this is truly an equivalence relation? I understand that reflexivity is true by definition of $\sim$ but I'm confused about we would proceed to show symmetry and transitivity.

Usually, I've seen equivalence relations as something like this:

Define $\sim$ on $\mathbb{R}$ as follows:

$x\sim y \iff x-y\in\mathbb{Z}$

I understand how to prove equivalence relations like the one above, by showing reflexivity, symmetry, and transitivity, but I'm stuck on how to proceed with the example from the topology textbook.

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Beneschan
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1 Answers1

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In fact, what you have provided is not an equivalence relation because it is not transitive: $(0.5,0)\sim(0,0.5)\sim(1,0.5)$ but $(0.5,0)\not\sim(1.0.5)$.

Probably what the textbook was referring to was the equivalence relation generated by those relations i.e. two points are in the same equivalence class iff you can prove that they are assuming the relations provided, assuming reflexivity, assuming symmetry, and assuming transitivity. For example, $(0.5,0)\sim(1.0.5)$ by applying transitivity to $(0.5,0)\sim(0,0.5)\sim(1,0.5)$.

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