Let $A$ and $B$ be two non empty sets of real numbers such that $A \cup B = (0,1)$ then this implies that $ \inf(A)\inf(B) = 0 $ and if the claim is not true then give an example to support your answer!
My approach was to prove by contradiction by assuming that $\inf(A)= a$ and $\inf (B)=b$ where both are greater than zero. So as the interval is uncountable if both were to be non zero then elements from $(0,a)$ would be missing from the union if $b \gt a \gt 0$ as $\inf(A \cup B) = \min(\inf(A),\inf(B))$ which contradicts the fact that $\inf(A \cup B)=0$
A question from real analysis
real-analysissupremum-and-infimum
Best Answer
Your idea is good, but you're too fast.
Let $a=\inf A$ and $b=\inf B$. Let $c=\min(a,b)/2$. If $c>0$, then either $c\in A$ or $c\in B$; on the other hand, $0<c<a$ implies $c\notin A$ and $0<c<b$ implies $c\notin B$. Therefore $c=0$, so $\min(a,b)=0$.