$\forall{a\in A,\, b\in B} \mid a\le b$ Prove that: $\sup A \le \inf B$

Question

Let A and B be two non-empty bounded subsets of $\mathbb{R}$.

$\forall{a\in A,\, b\in B} \mid a\le b$

Prove that: $\sup A \le \inf B$

My solution goes as follows:

Suppose $\sup A \gt \inf B$:

$\forall {a\in A,b\in B}\,\,\exists{\varepsilon>0}\mid (a + \varepsilon \gt \sup A)\land (b - \varepsilon \lt \inf B) $.

Therefore, $\space a + \varepsilon \gt b - \varepsilon \space\space \rightarrow \space\space a + 2\varepsilon \gt b$

Why did I not get a contradiction?

Answer 1

Firstly, your approach is correct but you seem to be a bit jumbled in all the technicalities. I think the proof that wj32 gave is more natural, but here is a proof in the same flavour as yours.

For all $\epsilon > 0$ there exists $a\in A$ and $b\in B$ such that $$a + \epsilon \ge \sup A,\ \ \ b-\epsilon \le \inf B$$ So far this is exactly what you stated, except without the logical symbols (you had the quantifiers backwards). Now suppose for the sake of contradiction that $\sup A > \inf B$. Then from the above inequalities we would have $$a + \epsilon \ge \sup A > \inf B \ge b-\epsilon$$ So for all $\epsilon > 0$ there exists some $a$ and some $b$ such that $$a > b - 2\epsilon$$ Since $\epsilon$ is arbitrarily small, this means that there exists $a$ and $b$ such that $a \ge b$ (this needs some formalizing). Now either there exists $a > b$ which contradicts our hypothesis, or we have $a=b$ where in this case we simply have $\sup A = \inf B$ contrary to our assumption.

Answer 2

To get a contradiction you need to explore interesting values: when dealing with proof with $\epsilon$ I find it helpful to plug in some real numbers and try to get an example working first:

Take $A=[0,1)$ and $B=[1,2]$. Hence we have $\sup{A}=1$, $\inf{B}=1$. In your proof, choose $a=0.5$, $b=2$. Then you would need to choose $\epsilon=2$ for example (you need $\epsilon>1$ at least). A telling sign that something is not going in the right direction here is that epsilon has a big value, you should tipically think of epsilon as a very small number.

Since it seems an homework question I won't say more for now, try to take my specific example and see what would happen if your assumption was true (i.e. for ex. with $\sup{A}=1.5$) and using the definition of $$\sup{A} = \max_{x}\{x | \forall \epsilon > 0 \quad\exists a\in A: |x-a|<\epsilon \} $$

If you still can't solve it I'll post a complete solution on request.

Answer 3

Why did I not get a contradiction?

What is the main argument or idea in your proof? Does it make sense?

Here are two proofs for you to complete:

Proof 1. Let $b \in B$; we have $a \le b$ for all $a \in A$. Since $b$ is an $____$ of $___$, $\sup A \le b$. But this is true for all $b$, so $\sup A$ is a $____$ of $___$ and therefore $\sup A \le \inf B$.

Proof 2. Suppose $\sup A > \inf B$ and let $\varepsilon = (\sup A - \inf B)/2$. Since $___$, there exists some $a \in A$ with $\sup A - \varepsilon < a \le \sup A$. Also, since $___$, there exists some $b \in B$ with $\inf B \le b < \inf B + \varepsilon$. This contradicts $___$.