# Is this the correct relation for a finite set?

Given a finite set S, let the relation

R = {(S1, S2) | |S1| < |S2|, S1, S2 ⊆ S}.

Show whether or not R is reflexive, symmetric, antisymmetric or transitive.

So if S = {1,2}

$$R = \{(\emptyset,\{1\}),\,(\emptyset,\{2\}),\,(\emptyset,\{1,2\}),\,(\{1\},\{1,2\}),\,(\{2\},\{1,2\})\}$$

and going off of this

Reflexive- No because S1 does not equal S2

Symmetric- No because if you swap $$(\emptyset,\{2\})$$ for example, the relation doesn't hold true

Transitive- No because $$\{(\emptyset,\{1\}),\,(\emptyset,\{2\})$$ and doesn't have $$\{(\{1\},\{2\})$$

Anti-Symmetric-No also because it doesn't satisfy aRb and bRa being true, a = b.

I think I've done this correctly, but I dont have a strong grasp on these relations yet.

• This relation is neither reflexive nor symmetric nor anti-symmetric. This relation is only transitive. – Dbchatto67 Apr 21 at 4:48

## 1 Answer

This relation is only transitive. Because if $$(S_1,S_2) \in R$$ and $$(S_2,S_3) \in R$$ then we have $$|S_1| < |S_2|\ \text {and}\ |S_2| < |S_3|.$$ But it follows that $$|S_1| < |S_3|.$$ So $$(S_1,S_3) \in R.$$

• Alright that makes sense, I see how i messed up! Just to be clear, first off is my set of relations correct? Secondly, is my reasoning for the others being false correct? – Brownie Apr 21 at 4:56
• Your defined relation is correct. Your reasoning for other parts (excluding transitivity) are perfectly fine. For concluding transitivity you cannot take $\{\varnothing, {1} \}, \{\varnothing,{2} \} \in R.$ Because $R$ is not symmetric. – Dbchatto67 Apr 21 at 4:58
• Because $$\{\varnothing ,a \} \in R \not\implies \{a,\varnothing \} \in R.$$ for $a=1,2.$ Hence for transitivity to hold we cannot say that $\{1,2 \} \in R.$ – Dbchatto67 Apr 21 at 5:05