# IF a set is bounded below, then its complement is not bounded below

Let $A \subset \mathbb{R}$ with $A$ bounded below. I want to show that $\mathbb{R} \setminus A$ is not bounded below.

Suppose not: that is suppose there is some $\alpha \in \mathbb{R}$ such that $\alpha \leq x$ for all $x \in \mathbb{R} \setminus A$. I dont see how can I arrive to a contradiction from here. I dont see also how to give a direct proof.

If you have a lower bound $b$, then every number smaller than $b$ is a lower bound as well, and is contained in $\Bbb{R} \setminus A=A^C$. So $$(-\infty, b) \subseteq A^C$$ is not bounded below (I hope this is obvious), neither is $A^C$.

• thanks very much. one question: is it possible to find a set $A \subset \mathbb{R}$ for with both $A$ and $A^C$ are both unbounded below and above ? – user203867 Apr 22 '15 at 7:53
• Of course. Take $A=\Bbb{Z}$, for example. – Crostul Apr 22 '15 at 7:55
• I know it looks obvious but how can we show that $\mathbb{Z}^C$ is unbounded below and above ? – user203867 Apr 22 '15 at 7:57
• by this logic, we can also take $A = \mathbb{Q}$ correct? – user203867 Apr 22 '15 at 7:57
• @Anonaki Yes. There are lots of examples. $\Bbb{Q}$ is another good one. – Crostul Apr 22 '15 at 8:01

You can pick $\beta$ with the same property with respect to $A$. Now have a look at $\min\{\alpha,\beta\}-1$.

• I dont understand so we know $\beta \leq a$ for all $a \in A$. but how does $\min( \alpha, \beta) - 1$ helps ? – user203867 Apr 22 '15 at 7:42
• Show that this number is neither contained in $A$ nor in $\mathbb R \setminus A$. This is a clear contradiction, isn't it? – MooS Apr 22 '15 at 7:42
• To be honest, I dont see the contradiction. Sorry I am very slow in math – user203867 Apr 22 '15 at 7:43
• If you are slow in math, why do you answer after a few seconds, that you do not see the contradiction? Give yourself some time.... – MooS Apr 22 '15 at 7:44