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.

The typesetting fix is good. Can you also describe what $\mathfrak{R}$ is? and how you compare sets $X$ that might be in your specified set. — TravisJ, Apr 03 '15 at 00:12
$\mathfrak{R}$ is the real numbers and I'm not sure what you're asking me about X. My professor said to take the whole thing as a family set if that helps. — ChemDude, Apr 03 '15 at 00:19

score 0 · Answer 1 · answered Apr 03 '15 at 00:39

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.

score 0 · Answer 2 · answered Apr 03 '15 at 19:24

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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

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

answered Apr 03 '15 at 19:24

Brian M. Scott

616,228

Ok here's how I started out, Take a y $\in$ $\mathscr{C}$ where y $\ge$ x+1. Clearly y $\subseteq$ $\Bbb R$ and y $\neq$ x and y satisfies $\forall x\forall y((x\in X\wedge y<x)\rightarrow y\in X$) but y is not a minimal element. So, $\mathscr{C}$ has no minimal elements. Is this correct? – ChemDude Apr 03 '15 at 21:02
@MathHelpNeeded: Members of $\mathscr{C}$ are subsets of $\Bbb R$, not real numbers, so if $y\in\mathscr{C}$ it makes no sense to write $y\ge x+1$. Also, the $X$ in your quantified expression is undefined, so the expression is not meaningful. Back up. Is the set $[x+1,\to)$ an element of the family $\mathscr{C}$? – Brian M. Scott Apr 03 '15 at 21:09
yes it is a member. I'm not understanding what you're doing? could you possibly give me the proof? – ChemDude Apr 03 '15 at 23:37
@MathHelpNeeded: Just finish the argument by showing that $[x+1,\to)\subsetneqq M$, contradicting the minimality of $M$. – Brian M. Scott Apr 04 '15 at 13:45

How do I prove this? (Relations Proof)

2 Answers2