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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.

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    $\begingroup$ This relation is neither reflexive nor symmetric nor anti-symmetric. This relation is only transitive. $\endgroup$ – Dbchatto67 Apr 21 at 4:48
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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 $(S_1,S_3) \in R.$

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  • $\begingroup$ 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? $\endgroup$ – Brownie Apr 21 at 4:56
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    $\begingroup$ 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. $\endgroup$ – Dbchatto67 Apr 21 at 4:58
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    $\begingroup$ Because $$\{\varnothing ,a \} \in R \not\implies \{a,\varnothing \} \in R.$$ for $a=1,2.$ Hence for transitivity to hold we cannot say that $\{1,2 \} \in R.$ $\endgroup$ – Dbchatto67 Apr 21 at 5:05

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