How do I prove this? (Relations Proof)

Question

So I can't seem to figure out how to prove this. Any help would be greatly appreciated. My professor said a contradiction would work but I don't see where I can make a contradiction.

Show that {X $\subseteq \Re | X \neq \varnothing$ and $\forall _x \forall _y ((x\in X \land x<y)\longrightarrow y\in X$}has no minimal element.

Answer 1

Well, it really depends on the quantity of set-theoretic details you require, but I'd try the following. Let

$\Gamma:=\{X\subseteq\mathbb{R}\ |\ X\neq\emptyset\wedge\forall x\forall y((x\in X\wedge y<x)\rightarrow y\in X)\}$

and $M\in\Gamma$ minimal, i.e. $\forall Y\in\Gamma(Y\subseteq M\rightarrow Y=M)$

Let's recall notations $$ [x,+\infty):=\{y\in\mathbb{R}\ |\ x\leq y\}\\ (x,+\infty):=\{y\in\mathbb{R}\ |\ x< y\} $$ Also, note that $\forall x\in\mathbb{R}\ (\ [x,+\infty)\in\Gamma\wedge(x,+\infty)\in\Gamma)$.

$M\in\Gamma\Rightarrow\exists x\in M\Rightarrow (x,+\infty)\subseteq[x,+\infty)\subseteq M$

But these sets are all in $\Gamma$, so by minimality $(x,+\infty)=M=[x,+\infty)$; absurd.

Answer 2

HINT: Let $\mathscr{C}$ be the given family of subsets of $\Bbb R$, and suppose that $M\in\mathscr{C}$ is minimal. $M\in\mathscr{C}$, so $M\ne\varnothing$, and we may choose $x\in M$. Now consider the set

$$[x+1,\to)=\{y\in\Bbb R:y\ge x+1\}\;.$$