I'm trying to show $\inf(A) = \inf(B)$, for $B$ contained in $A$ and $a+m\in B$, for all $a\in A$ and $m>0$.
I have shown that $\inf(A)\leq\inf(B)$, but I'm having a hard time showing that the $\inf(B)\leq \inf(A)$. So, I have shown that the greatest lower bound of $A$ will be the greatest lower bound of $B$, but I am having a hard time showing that the greatest lower bound of $B$ is also the greatest lower bound of $A$.
So, I've shown that $\inf(A)\leq\inf(B)$ by showing that $a\leq b$, but I can't seem to show $b\leq a$. Do I even need to show $b\leq a$ to prove $\inf(B) \leq \inf(A)$?
Here's what I tried and failed at :
$\ InfB\leq\ b = a+m$ for all $a∈A$.
So, $\ InfB\leq\ a+m =b$ for all $a∈A$.
So, $\ InfB\leq\ b-m = a$ for all $b∈B,a∈A$. (This part seems wrong from here an moving on)
So, $\ InfB\leq\ a$.
I'm trying to do a direct proof here, so is there a way to do this without doing a proof by contradiction?
Let's rephrase the problem slightly:
Let $A$ and $B$ be non-empty subsets of $\mathbb{R}$ that are bounded below. (This ensures that the infima $\text{inf}(A)$ and $\text{inf}(B)$ are well-defined real numbers.) Suppose that for each $a\in A$ and $m>0$, we have that $a+m\in B$. You are trying to prove that $\text{inf}(A)=\text{inf}(B)$ and have already done half of the proof by showing that $\text{inf}(A)\leq\text{inf}(B)$. All that remains is to show that $\text{inf}(A)\geq\text{inf}(B)$, which we prove by contradiction below:
Proof: Suppose that $\text{inf}(A)<\text{inf}(B)$. By the definition of the infimum, there exists $a\in A$ with $a<\text{inf}(B)$. Then for each $m\in\mathbb{R}$ with $0<m<\text{inf}(B)-a$, we have that $a+m\in B$, and since $\text{inf}(B)$ is (by definition) a lower bound for $B$, we have that $$a+m\geq\text{inf}(B)\implies m\geq\text{inf}(B)-a,$$ a clear contradiction. and we are done.