Question

if A be a subset of real number and B is real number then show that sup(b+a)=b+sup(A)

Answer 1

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Task. If $A$ be a subset of real number and $b$ is real number then show that $\sup(b+A)=b+\sup(A)$.

Proof. By definition

$x=\sup(A)$

if $x\geq a$ for every $a\in A$, and for every $\varepsilon>0$ there exists $a\in A$ such that

$a>x-\varepsilon.$

Also notice that

$b+A=\{b+a\mid a\in A\}.$

It suffices to prove that

$\sup(b+A)\leq b+\sup(A).$

Then applying this identity to $A^{\prime}=b+A$ and $b^{\prime}=-b$ we will obtain that

$\sup(b^{\prime}+A^{\prime})\leq b^{\prime}+\sup(A^{\prime}),$

that is

$\sup(-b+b+A)\leq-b+\sup(b+A)$

$\sup(A)\leq-b+\sup(b+A),$

$\sup(A)+b\leq\sup(b+A).$

Which will imply that

$\sup(A)+b=\sup(b+A).$

Let $x=\sup(A)$ and $y=\sup(b+A)$. We should prove that

$y\leq b+x.$

Suppose $y>b+x$. This means that there exist $\varepsilon>0$ such that

$y>y-\varepsilon>b+x.$

But then there exist $z\in b+A$ such that

$z>y-\varepsilon>b+x.$

Notice that $z$ has the fowm $z=b+a$ for some $a\in A$, whence

$z=b+a>b+x$

and so

$a>x$

which contradicts to the assumption that $x=\sup(A)\geq a$.

Hence $y\leq b+x$. And so

$\sup(A)+b=\sup(b+A).$