"Let $X \subset \mathbb{R}$ be nonempty and $a > 0$. Define
$$a + X = \{a+x: x \in X\}$$
Show that if $X$ is bounded above, then there exists $y \in a + X$ such that $y$ is an upper bound of $X$"
My approach would be to prove by contradiction:
"Suppose that $X$ is bounded and there is no $y \in a + X$ such that $(y > x)$ $ \forall x \in X$
But then $x + a \not > x \iff a \not > 0$
This is a contradiction, since by definition $a > 0$"
However the "hint" in the question suggests that we should prove this by contrapositive. How would I approach this?
The contrapositive of the statement of the theorem is the natural place to begin.
Let a >0 $\in\mathbb R$. If there is no y$\in$a + X $\subseteq\mathbb R$ such that y is an upper bound of X, then X cannot be bounded above.
Assume X $\subseteq\mathbb R \neq \emptyset$. Let y$\in a + X$. Assume y cannot be an upper bound for X.Also assume X is bounded above. Since X $\subseteq\mathbb R \neq \emptyset$ and X is bdd above, there exists z$\in \mathbb R$ such that z is the least upper bound of X. This means for every x$\in$X where l$\in\mathbb R$ is an upper bound of x,$1\geq z$ and $z\geq x$. Since a>0, y = a+ x >0.Since z is the least upper bound of X,z$\geq$ x. This means z-x$\geq 0$.If z -x = $0$, then z =x. Then y>z for every y$\in a+X $ But then y is an upper bound of X and we have a contradiction. If z-x >$0$, then there exists l>$0\in\mathbb R$ such that z= l+x. But then since l>$0$, by definition, z$\in$ a+X and again we have a contradiction since z is an upper bound of X! So we're forced to conclude X is not bounded above and we're done. Q.E.D.
P.S. Minor error in the proof. Just corrected it, glad I caught it. Would have been humiliating......