# How to prove this fact about the discrete closure? [closed]

The content is given two relationships: R₁ and R₂ prove that s(R₁ ∩ R₂)=s(R₁) ∩ s(R₂)

My teacher has taught us the UNION versions in class, and I figure it's easy. Also I have already finished the other two intersection versions( transitive and reflexive closure), but the symmetric practice stucks me a lot,I found if expand one side directly ,it seems impossible to testify that they are equal.

## closed as unclear what you're asking by Jean-Claude Arbaut, Mike Earnest, Joshua Mundinger, Lord Shark the Unknown, Lee David Chung LinApr 20 at 1:17

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Let’s prove, that $$S(R_1 \cap R_2) \subset S(R_1) \cap S(R_2)$$:
If $$(a, b) \in S(R_1 \cap R_2)$$, that means that either $$(a, b) \in R_1 \cap R_2$$ or $$(b, a) \in R_1 \cap R_2$$. That means that either $$(a, b)$$ or $$(b, a)$$ is in both $$R_1$$ and $$R_2$$ at the same time. And that can be rephrased as $$(a, b)$$ being in both $$S(R_1)$$ and $$S(R_2)$$ at the same time, which is exactly $$(a, b) \in S(R_1) \cap S(R_2)$$.
However, as pointed in the comments, the converse does not hold: If $$R_1 = \{(a, b)\}$$ and $$R_2 = \{(b, a)\}$$, then $$S(R_1) \cap S(R_2) = \{(a, b), (b, a)\}$$, but $$S(R_1 \cap R_2) = \emptyset$$
• Hmm. I thought the other containment is the difficult one. In fact, isn't it false? If $R_1=\{(a,b)\}$ and $R_2=\{(b,a)\}$ then $S(R_1)\cap S(R_2)=\{(a,b),(b,a)\}$ but $S(R_1\cap R_2)$ is empty unless I misunderstood something. – Jyrki Lahtonen Apr 19 at 19:12