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Let $A \subseteq \mathbb { R }$ be a nonempty set which is bounded from below in $\mathbb { R }$ . Given a fixed number $k \in \mathbb { R } ,$ show that the set $k + A : = \{ k + a : a \in A \}$ is bounded from below and that $\inf ( k + A ) = k + \inf ( A ).$

Where do I begin with this proof? Do I need to show $\inf(k+A)\leq k + \inf(A)$ and $\inf(k+A)\geq k + \inf(A)$?

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    $\begingroup$ Remember in general that to show that $\inf(S) = m$, it suffices to show two things: 1) that $x\geq m$ for all $x\in S$, and 2) if $m' > m$, then there exists $x\in S$ with $x < m'$. So to show that $\inf(k+A) = k + \inf(A)$, you can try and show that 1) $x\geq k + \inf(A)$ for all $x\in k + A$, and 2) if $m' > k + \inf(A)$, then there exists $x \in k + A$ with $x < m'$. $\endgroup$ – Minus One-Twelfth Feb 23 at 0:17
  • $\begingroup$ Related 1 and 2. $\endgroup$ – Git Gud Feb 23 at 0:49
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    $\begingroup$ It might help to name $s=\inf A$, write down what that means about $s$, then try to make similar statements from those but in terms of $s+k$. For example, if $s\leq a$, then $k+s\leq k+a$. $\endgroup$ – MPW Feb 23 at 0:50
  • $\begingroup$ Let $\alpha=\inf A.$ Clearly $k + a \geq k + \alpha$ while there exists $a_n$ with $a_n < \alpha+\frac{1}{n}$ hence $a_n+k < \alpha+\frac{1}{n}.$ $\endgroup$ – Will M. Feb 23 at 1:06
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User @MinusOne-Twelfth has already given the outline of the proof. I will show that $\sup ( k + A ) = k + \sup ( A )$(Assuming $A \subseteq \mathbb { R }$ is a non-empty subset bounded from above). Can you prove it for $\inf (k+A) $?


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

Proof: Recall that every non-empty set bounded from above has supremum. So $\sup (A) $ exists and $a\leq \sup ( A )$ for all $a\in A $. So $k+a\leq k+\sup ( A )$ for all $a\in A $. Hence $k+\sup ( A )$ is an upper bound for the set $k+A $. In particular, $\sup ( k + A ) \leq k + \sup ( A )\tag1.$

Now $k+a\leq \sup ( k + A )$ for all $a\in A .$ This implies that $a\leq \sup ( k + A )-k$ for all $a\in A .$ So $\sup ( k + A )-k$ is an upper bound for the set $A .$ Hence $\sup(A)\leq \sup ( k + A )-k\iff k+\sup (A)\leq\sup ( k + A )\tag2.$ From $(1) $ and $(2)$ we can conclude $\sup ( k + A ) = k + \sup ( A ).$ $\blacksquare $

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The idea is to "back off the inf's" and use a "since $\epsilon$ is arbitrary" argument. More precisely,

$1).\ A$ is bounded below, which implies that $A+k$ is also. This means that each has a finite greatest lower bound.

$2).\ $ Set $\alpha=\inf A$ and $\beta=\inf (A+k)$. You want to show that $\beta=\alpha+k.$ As you rightly point out, if we can show $\beta\le\alpha+k$ and $\beta\ge\alpha+k,$ we will have proved the claim.

Let $\epsilon>0$.

$3).\ $ There is an $a\in A$ such that $a<\alpha+\epsilon.$ Then, $a+k\in A+k$ and so $\beta\le a+k<\alpha+k+\epsilon.$ Since $\epsilon>0$ is arbitrary, we get $\beta\le\alpha+k$.

$4).$ There is a $b\in A+k$ such that $b\le \beta+\epsilon.$ But $b=a+k$ for some $a\in A$ (by definition of the set $A+k$) and of course $\alpha\le a.$ Therefore, $\alpha+k\le a+k=b< \beta+\epsilon$ and so by the same reasoning as in $(3)$, we get $\alpha +k\le \beta.$

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