This is an exercise from Velleman's "How To Prove It". I have struggled with this problem for a while, so I just want to make sure that it is correct.
Show that $\{X \subseteq \mathbb{R} | X \neq \emptyset \wedge \forall x \forall y((x \in X \wedge x < y) \rightarrow y \in X)\}$ has no minimal elements.
Let $$L = \{X \subseteq \mathbb{R} | X \neq \emptyset \wedge \forall x \forall y((x \in X \wedge x < y) \rightarrow y \in X)\}$$
To say that $L$ has some minimal element (under the subset partial order), we could write $$\exists X \in L \forall Y \in L (Y \subseteq X \rightarrow Y = X)$$
We want to show that $L$ does not have a minimal element, so we need to prove
$$\forall X \in L \exists Y \in L ( Y \subseteq X \wedge Y \neq X)$$
Proof: Let $X \in L$ be arbitrary. Since $X \in L$, $X \neq \emptyset$ and $\forall x \forall y ((x \in X \wedge x < y) \rightarrow y \in X)$. Since $X \neq \emptyset$, we can choose some $a \in X$. Then, in particular, $\forall y (a < y \rightarrow y \in X)$. We can then say that $X = \{y \in \mathbb{R} | y \geq a\}$. But now consider the set $X' = \{y \in \mathbb{R} | y \geq a+1\}$. Clearly, $X' \subseteq X$, and $X' \neq \emptyset$ since $a + 1 \in X'$. Now let $x$ and $y$ be arbitrary such that $x \in X'$ and $x < y$. Since $x \in X'$, $x \geq a + 1$. Since $y > x$, $y > a + 1$. It follows that $y \in X'$. Thus, $X' \in L$. We have found a set $X' \in L$ such that $ X' \subseteq X$ but $X' \neq X$. Since $X \in L$ was arbitrary, $L$ has no minimal elements. $\square$
Best Answer
Your proof goes astray when you say that $X=\{y\in\Bbb R:y\ge a\}$: that would be true if $a$ were the minimum element of $X$, but you have no reason to think that that is the case. Thus, all you can say is that $X\supseteq\{y\in\Bbb R:y\ge a\}$. Fortunately, this error isn’t fatal, since you can still argue that $X'\subsetneqq\{y\in\Bbb R:y\ge a\}\subseteq X$ and hence — since you correctly showed that $X'\in L$ that $X$ is not minimal in $L$.
By the way, we can describe $L$ completely: the members of $L$ are $\Bbb R$, all open rays $(x,\to)$ for $x\in\Bbb R$, and all closed rays $[x,\to)$ for $x\in\Bbb R$. (You are probably more familiar with the notation $(x,\infty)$ and $[x,\infty)$.) These are precisely the non-empty subsets $X$ of $\Bbb R$ that have the property that
$$X=\bigcup_{x\in X}[x,\to)\,;$$
this equation is equivalent to the definition that you were given.