# Show $\inf(k+A)= k + \inf(A)$

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 $$\inf(k+A)\geq k + \inf(A)$$?

• 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 $x < m'$. – Minus One-Twelfth Feb 23 at 0:17
• Related 1 and 2. – Git Gud Feb 23 at 0:49
• 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 $k+s\leq k+a$. – MPW Feb 23 at 0:50
• Let $\alpha=\inf A.$ Clearly $k + a \geq k + \alpha$ while there exists $a_n$ with $a_n < \alpha+\frac{1}{n}$ hence $a_n+k < \alpha+\frac{1}{n}.$ – Will M. Feb 23 at 1:06

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

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 $$\alpha +k\le \beta.$$