Question

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

Answer 1

.

Task. If A is subset of real number and $b$ is a real number, then show that

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

Proof. By definition

$x=\inf(A)$

if $x\leq a$ 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

$\inf(b+A)\geq b+\inf(A).$

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

$\inf(b^{\prime}+A^{\prime})\geq b^{\prime}+\inf(A^{\prime}),$

that is

$\inf(-b+b+A)\geq-b+\inf(b+A)$

$\inf(A)\geq-b+\inf(b+A),$

$\inf(A)+b\geq\inf(b+A).$

Which will imply that

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

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

$y\geq b+x.$

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

$b+a<b+x$

and so

$a<x$

which contradicts to the assumption that $x=\inf(A)\leq a$.

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

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