Prob. 7, Sec. 5, in G.F. Simmon's INTRO. TO TOPOLOGY & MODERN ANALYSIS: Equivalence relation iff reflexive and circular iff reflexive and triangular Here is Prob. 7, Sec. 5, in the book Introduction to Topology and Modern Analysis by George F. Simmons:

Let $X$ be a non-empty set. A relation $\sim$ in $X$ is called circular if $x \sim y$ and $y \sim z$ $\implies z \sim x$, and triangular if $x \sim y$ and $x \sim z$ $\implies y \sim z$. Prove that a relation in $X$ is an equivalence relation $\iff$ it is reflexive and circular $\iff$ it is reflexive and triangular.

My Attempt:

Suppose that $\sim$ is an equivalence relation in $X$. Then the following three statements hold:
(1) $x \sim x$ for every $x \in X$.
(2) For any $x, y \in X$, if $x \sim y$, then $y \sim x$ also.
(3) For any $x, y, z \in X$, if $x \sim y$ and $y \sim z$, then $x \sim z$ also.
Let $x, y, z \in X$ such that $x \sim y$ and $y \sim z$. Then by (3) we have $x \sim z$, and hence by (2) we have $z \sim x$, thus showing that relation $\sim$ in $X$, which is already reflexive, is circular as well.

Am I right?

Now suppose that relation $\sim$ in $X$ is reflexive and circular. Then the following two statements hold:
(1) For any $x \in X$, we have $x \sim x$.
(2) For any $x, y, z \in X$, if $x \sim y$ and $y \sim z$, then $z \sim x$.
Let $x, y, z \in X$, and suppose that $x \sim y$ and $x \sim z$.
But we also have $x \sim x$.
Now as $x \sim x$ and $x \sim y$, so we also have $y \sim x$ as well.
Thus we have $y \sim x$ and $x \sim z$. Therefore we also have $z \sim y$.
Finally, as we have $z \sim z$ and $z \sim y$, so we also have $y \sim z$.
Thus we have shown that, for any elements $x, y, z \in X$, if $x \sim y$ and $x \sim z$, then $y \sim z$ also.
Therefore the relation $\sim$ on $X$, which is already reflexive, is also triangular.

Am I right?

Lastly, suppose that relation $\sim$ on $X$ is reflexive and triangular. Then the following two statements hold:
(1) For any element $x \in X$, we have $x \sim x$.
(2) For any elements $x, y, z \in X$, if $x \sim y$ and $x \sim z$, then we also have $y \sim z$.
Let $x, y, z \in X$.
First suppose that $x \sim y$. Then as $x \sim y$ and $x \sim x$, so we also have $y \sim x$. Thus relation $\sim$ is symmetric.
Next suppose that $x \sim y$ and $y \sim z$. Then we have $y \sim x$ and $y \sim z$, and therefore $x \sim z$. Thus relation $\sim$ is transitive.
Therefore relation $\sim$ on $X$, which was already reflexive, is both symmetric and transitive as well. Hence $\sim$ is an equivalence relation on $X$.

Am I right?
Are all of the above proofs correct and clear enough? If not, then where are the problems?
 A: It all looks correct! It could be clearer, though. Let me go through and make some suggestions.

Suppose that $\sim$ is an equivalence relation in $X$. Then the following three statements hold:
(1) $x \sim x$ for every $x \in X$.
(2) For any $x, y \in X$, if $x \sim y$, then $y \sim x$ also.
(3) For any $x, y, z \in X$, if $x \sim y$ and $y \sim z$, then $x \sim z$ also.
Let $x, y, z \in X$ such that $x \sim y$ and $y \sim z$. Then by (3) we have $x \sim z$, and hence by (2) we have $z \sim x$, thus showing that relation $\sim$ in $X$, which is already reflexive, is circular as well.

You've got exactly the right idea. I'd go about it this way:

*

*Suppose that $\sim$ is an equivalence relation on $X.$ Then $\sim$ is reflexive by definition. We show that it is circular. Take any $x,y,z\in X$ such that $x\sim y$ and $y\sim z.$ By transitivity, we have $x\sim z,$ so we have $z\sim x$ by symmetry. Thus, $\sim$ is circular.

There's nothing wrong with your approach, but it can be nice to announce your goals, and to use the terminology over symbols.

Now suppose that relation $\sim$ in $X$ is reflexive and circular. Then the following two statements hold:
(1) For any $x \in X$, we have $x \sim x$.
(2) For any $x, y, z \in X$, if $x \sim y$ and $y \sim z$, then $z \sim x$.
Let $x, y, z \in X$, and suppose that $x \sim y$ and $x \sim z$.
But we also have $x \sim x$.
Now as $x \sim x$ and $x \sim y$, so we also have $y \sim x$ as well.
Thus we have $y \sim x$ and $x \sim z$. Therefore we also have $z \sim y$.
Finally, as we have $z \sim z$ and $z \sim y$, so we also have $y \sim z$.
Thus we have shown that, for any elements $x, y, z \in X$, if $x \sim y$ and $x \sim z$, then $y \sim z$ also.
Therefore the relation $\sim$ on $X$, which is already reflexive, is also triangular.

Again, you're going in exactly the right direction, but I'd do it a bit differently.

*

*Suppose that $\sim$ is a relation on $X$ that is reflexive and circular. We show that it is triangular. Take $x,y,z\in X$ such that $x\sim y$ and $x\sim z.$ By reflexivity, we have $x\sim x,$ so since $x\sim y,$ then $y\sim x$ by circularity. Since $y\sim x$ and $x\sim z,$ then $z\sim y$ by circularity. By reflexivity, we have $z\sim z,$ so since $z\sim y,$ then $y\sim z$ by circularity. Thus, $\sim$ is triangular.


Lastly, suppose that relation $\sim$ on $X$ is reflexive and triangular. Then the following two statements hold:
(1) For any element $x \in X$, we have $x \sim x$.
(2) For any elements $x, y, z \in X$, if $x \sim y$ and $x \sim z$, then we also have $y \sim z$.
Let $x, y, z \in X$.
First suppose that $x \sim y$. Then as $x \sim y$ and $x \sim x$, so we also have $y \sim x$. Thus relation $\sim$ is symmetric.
Next suppose that $x \sim y$ and $y \sim z$. Then we have $y \sim x$ and $y \sim z$, and therefore $x \sim z$. Thus relation $\sim$ is transitive.
Therefore relation $\sim$ on $X$, which was already reflexive, is both symmetric and transitive as well. Hence $\sim$ is an equivalence relation on $X$.

I'd probably do this differently, too.

*

*Suppose that $\sim$ is a relation on $X$ that is reflexive and triangular. We show that $\sim$ is symmetric and transitive, so that $X$ is an equivalence relation on $X.$


*To show symmetry, take $x,y\in X$ such that $x\sim y.$ By reflexivity, we also have $x\sim x,$ so by triangularity we have $y\sim x,$ as desired.


*To show transitivity, take $x,y,z\in X$ such that $x\sim y$ and $y\sim z.$ By symmetry, we have $y\sim x,$ so since $y\sim z,$ then by triangularity we have $x\sim z,$ as desired.
