Showing Parallelism is an equivalence relation in $\Bbb R ^2$

Let $L$ denote the set of all lines in the Euclidean plane $\Bbb R^2$. We say that two lines $L$ and $M$ are parallel if either $L = M$ or $L$ and $M$ have no points in common.

Using Euclid’s Parallel Postulate prove that parallelism is an equivalence relation on the set $L$.

I need some help determining if I have the right idea about this.

$L=L$ which means that $L$ is parallel to itself. So this is reflexive.

Suppose we have $L || M$. Then Either $L=M$ or $L$ and $M$ share no points. So Then $M=L$ or $M$ and $L$ share no points. Then $M||L$. So this is symmetric.

Suppose we have $L || M$ and $M || N$. So then either $L=M$ or $L$ and $M$ share no points and either $M=N$ or $M$ and $N$ share no points. So we have that either $L=N$ or $L$ and $N$ share no points. So $L || N$. So then this is transitive.

So from all of this, parallelism is an equivalence relation. I’m not sure if this is correct or not. Any help would be appreciated.

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Your reasoning looks good. It might help to note explicitly that you are using symmetry of these two other relations, “$=$” and “share no points”, to conclude, for example:

$$L = M \implies M = L$$


$$\text{$L$ and $M$ share no points} \implies \text{$M$ and $L$ share no points}$$

ditto with reflexivity, transitivity.

In fact, these relations are both equivalence relations, and your proof is an expression of the fact that you can form a new equivalence relation from two old equivalence relations by joining them together with “or”.

But that’s a tangent. The primary benefit in being really pedantic and explicit about what facts you are using in these sorts of proofs, IMO, is the confidence boost.