Prove that symmetric closure of R $h_{sym}(R) = R \cup R^T$

Question

I have a question about proving statements of the form in which the given is the union of two sets. Well first of all let me just write how I tried to prove the statement:

To show that $h_{sym}(R) = R \cup R^T$, you can show that $h_{sym}(R) \subseteq R \cup R^T$ and $R \cup R^T \subseteq h_{sym}(R)$.

Assume $(x,y) \in h_{sym}(R)$. Since by definition $h_{sym} R = \bigcap \{S \subseteq A \times A : $ S is symmetric and $R \subseteq S \}$, it follows that $R\subseteq h_{sym}(R)$.

Now my question is, is this sufficient to prove that $(x,y)\in R \cup R^T$? Basically what $R \cup R^T$ means is that for any elements $(x,y)$, $(x,y)\in R \vee (x,y)\in R^T$, so it should be sufficient to prove that (x,y) is an element of one of the sets of the union, correct?

For the other direction:

Assume $(x,y)\in R\cup R^T$, then there are two cases to be looked at:

Case 1: Let $(x,y)\in R$, since $R\subseteq h_{sym}(R)$, it follows that $(x,y)\in h_{sym}(R)$.

Case 2: Let $(x,y)\in R^T$, since by definition $R\subseteq h_{sym}(R)$, it follows that $(x,y)\in R\cap R^T$, since $h_{sym}(R)$ is symmetric, it follows that $(x,y)\in h_{sym}(R)$.

As you can probably tell I'm still new to writing proofs and would very much appreciate an answer to my question and a brief feedback on whether the proof is correct (I have my doubts about Case 2 being correct) and in general what could be done better.

Answer 1

Your argument that $h_{sym}(R)\subseteq R\cup R^T$ simply doesn’t work. This is actually the harder direction. Here’s one way to do it.

Let $\langle x,y\rangle\in h_{sym}(R)=\bigcap\{S\subseteq A\times A:S\text{ is symmetric and }R\subseteq S\}$. Suppose that $\langle x,y\rangle\notin R\cup R^T$; then $\langle x,y\rangle\notin R$, and $\langle x,y\rangle\notin R^T$, so $\langle y,x\rangle\notin R$. Let $$T=h_{sym}(R)\setminus\{\langle x,y\rangle,\langle y,x\rangle\}\;.$$ Now get a contradiction by showing that $T$ is a symmetric relation on $A$, and $$R\subseteq T\subsetneqq h_{sym}(R)\;.$$ Conclude that $\langle x,y\rangle\in R\cup R^T$ after all.

However, there’s an easier way, using the fact that if $\mathscr{S}$ is any collection of sets, and $S\in\mathscr{S}$, then $S\supseteq\bigcap\mathscr{S}$:

Observe that $R\cup R^T$ is symmetric and contains $R$, so that $$R\cup R^T\in\{S\subseteq A\times A:S\text{ is symmetric and }R\subseteq S\}\;;$$ this immediately implies that $$R\cup R^T\supseteq\bigcap\{S\subseteq A\times A:S\text{ is symmetric and }R\subseteq S\}\;.$$

The second case of your argument for the opposite inclusion is also flawed. You have $\langle x,y\rangle\in R^T$, and you want to conclude that $\langle x,y\rangle\in h_{sym}(R)$. The fact that $R\subseteq h_{sym}(R)$ does not imply that $\langle x,y\rangle\in R$, which is what you’d need in order to justify your claim that $\langle x,y\rangle\in R\cap R^T$. However, if $\langle x,y\rangle\in R^T$, then $\langle y,x\rangle\in R\subseteq h_{sym}(R)$, and $h_{sym}(R)$ is symmetric, so $\langle x,y\rangle\in h_{sym}(R)$, as desired.