Prove that $\sup A+x=\sup(A+x)$

Question

Suppose $b$ is an upper bound for $A$.

That is, $y \le b$ for all $y \in A$.

Then, $x+y \le x+b$ for all $y \in A$, and so $x+b$ is an upper bound for $x+A$.

In particular, if $b = \sup A$, then $\sup(x+A) \le x+b = x+\sup(A)$.

The other direction is similar.

I don't understand the forth sentence. Why if $b=\sup(A)$, $x+A$ is transformed into $\sup(x+A)$?

Answer 1

You know that, for every upper bound $b$ of $A$, you have that $x+b$ is an upper bound for $x+A$. In particular, $x+\sup A$ is an upper bound for $x+A$.

Therefore, by definition of supremum, $\sup(x+A)\le x+\sup A$, because $\sup(x+A)$ is the least upper bound.

This is a general fact: if $B$ is any upper bounded set and $c$ is an upper bound of $B$, then $\sup B\le c$. Here $B=x+A$ and $c=x+\sup A$.