Show that $A + B$ is bounded below and that $\inf(A + B) = \inf(A) + \inf(B)$

Question

Suppose that $A$ and $B$ are non-empty subsets of $\textbf{R}$ which are bounded below. Let $A + B = \{x\in\textbf{R}\mid (x = a + b)\wedge(a\in A)\wedge(b\in B)\}$. Show that $A + B$ is bounded below and that $\inf(A + B) = \inf(A) + \inf(B)$. What about products?

MY ATTEMPT

Consult here.

Answer 1

Since $L_{1} + L_{2}$ is a lower bound, we must prove that $L = L_{1} + L_{2}$.

It suffices to show that $L \ge L_1 + L_2$, in fact; you get $L \le L_1 + L_2$ for free.

Given $a\in A$, there exists $\delta_{1}$ such that $L_{1} \leq a < L_{1} + \delta$. Similarly, given $b\in B$, there exists $\delta_{2}$ such that $L_{2} \leq b < L_{2} + \delta_{2}$. In other words, given $a + b\in A + B$, there exists $\delta_{1} + \delta_{2}$ such that $L_{1} + L_{2} \leq a + b < \delta_{1} + \delta_{2}$. This means that $L_{1} + L_{2}$ is the infimum of $A + B$.

You seem to have gotten this a little backwards. You should be using the fact that, because $L_1$ is the infimum of $A$, for all $\delta_1 > 0$, there exists an $a \in A$ such that $L_1 \le a < L_1 + \delta_1$. This is equivalent to $L_1$ being the infimum of $A$. What you wrote is fairly obviously true when $L_1$ is a lower bound to $A$, even if it's not the supremum (simply choose $\delta_1$ to be huge and positive). The conclusion, similarly, does not track either, as your $\delta$s could be very large.

Instead, keep it tight. Show that, for all $\varepsilon > 0$, there exists some $x \in A + B$ such that $$L_1 + L_2 \le x < L_1 + L_2 + \varepsilon.$$ Note, this is not for some $\varepsilon > 0$ (which is trivially true), but this is for all $\varepsilon > 0$. This means that $L_1 + L_2 + \varepsilon$ is not a lower bound to $A + B$, regardless of how small $\varepsilon > 0$ is, hence $L_1 + L_2$ is the infimum of $A + B$.

To do this, you have to construct $x$. Try using the fact that $L_1 + \varepsilon/2$ and $L_2 + \varepsilon/2$ are not, respectively lower bounds for $A$ and $B$.

As to the product $AB = \{x\in\textbf{R} \mid (x = ab)\wedge(a\in A)\wedge(b\in B)\}$, the result $\inf(AB) = \inf(A)\inf(B)$ holds, but we have to consider four cases according to the signs of the values $\inf(A)$ and $\inf(B)$. Let us deal with the case in which $\inf(A) > 0$ and $\inf(B) > 0$.

If we let $\inf(A) = L_{1}$ and $\inf(B) = L_{2}$, then given $a\in A$ and $b\in B$, we have that $a \geq L_{1} > 0$ and $b\geq L_{2} > 0$, from whence we conclude that $ab \geq L_{1}L_{2} > 0$. Thus we have to prove that $L = L_{1}L_{2}$ is the infimum indeed. In fact, given $\delta_{1} > 0$ and $\delta_{2} > 0$, there exist $a\in A$ and $b\in B$ such that $0 < L_{1} \leq a < L_{1} + \delta_{1}$ and $0 < L_{2} \leq b < L_{2} + \delta_{2}$. Consequently, given $\delta = L_{1}\delta_{2} + L_{2}\delta_{1} + \delta_{1}\delta_{2}$, there is $ab\in AB$, such that we have $0 < L_{1}L_{2} \leq ab < L_{1}L_{2} + \delta$, from whence we conclude that $L = L_{1}L_{2} = \inf(AB)$.

There's a similar issue here to the summation proof. You really need to start with arbitrary $\delta > 0$, and construct $a$ and $b$ from there.

That said, you're actually supposed to come up with a counterexample! Your assumption that $L_1, L_2 > 0$ will indeed make this thing work, but removing this assumption can easily cause this to no longer be true. See if you can come up with a counterexample.