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I do not know how to start. I tried to use property of transitive and reflexive to get $bRa$, but i cannot connect those together. Any hint on how to prove this?

B.LIANG
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2 Answers2

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Try it out on order $(\mathbb R,\leq)$.

Then you will find that no such element $b\in\mathbb R$ exists.

Simply because $a:=b-1\leq b$ and $a\neq b$ for every $b\in\mathbb R$.

drhab
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After question edit

If $|S|=1$ the statement is obvious. Now let's do full induction.

Suppose the statement holds for every nonempty partially ordered set with less than $|S|$ elements.

Let $a\in S$ and consider $T=\{x\in S:xRa, x\ne a\}$. If $T$ is empty, you are done. Otherwise, $|T|<|S|$ and so the induction hypothesis apply.

Now finish up the argument.

Original answer

You are asking whether any infinite partially ordered set has a minimal element.

Since total orders are partial orders, the property would imply that every infinite totally ordered set has a minimum. Now, can you find a counterexample?

egreg
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  • I cannot. But how would you prove that infinite partially ordered set has a minimal value. – B.LIANG Feb 16 '18 at 18:13
  • @B.LIANG You can't prove a false statement. Take the integers, for instance. – egreg Feb 16 '18 at 18:16
  • Sorry, I actually want to focus on finite set. That's a typo in the question... – B.LIANG Feb 16 '18 at 18:20
  • @B.LIANG OK, answer edited – egreg Feb 16 '18 at 18:38
  • Thank you. Solve that with proof by contradiction. – B.LIANG Feb 16 '18 at 23:44
  • @B.LIANG No contradiction: $T$ inherits the partial order and has less elements than $S$; by the induction hypothesis, it has a minimal element. $m$; suppose $b\in S$ and $bRm$; then $bRa$ by transitivity and $b\ne a$, otherwise $m=a$; thus $b\in T$ and therefore $b=m$ by minimality. – egreg Feb 16 '18 at 23:47