So my next question is again about uniform continuity. Can you give me hints, or (better) give the solution of the following exercise? Thank you very much 🙂
Given two subsets $A$ and $B$ of $\mathbb R$ with $A$ bounded from above (i.e., having an upper bound) and $B$ bounded from below (i.e., having a lower bound), where
$\sup A = \inf B$ and $\sup A \in A\cap B$(1) Prove that $A\cap B = \{\sup A\}.$
Now, take $A$ and $B$ as above. Let $f : \mathbb R \rightarrow \mathbb R$ and assume that $f$ is uniformly continuous on $A$ and on $B$.
(2) Prove that f is uniformly continuous on $A \cup B.$
My try:
(1) Let $x \in A\cap B$. Because $\sup A = \inf B,$ $\inf B \le x \le \sup A$ implies $x = \sup A$. I chose $x$ arbitrary, so $ A\cap B = \sup A$
(2) ???
Best Answer
The definition says $\forall\varepsilon>0\ \exists\delta>0\cdots\cdots\cdots\cdots$. To prove that a function is uniformly continuous, you need to find $\delta$ as a function of $\varepsilon$ and prove that it's small enough. You know you've got $\delta_1$ that's small enough on one set and $\delta_2$ that's small enough on the other. Which one is smaller might depend on $\varepsilon$. However $\min\{\delta_1,\delta_2\}$ will be small enough on both sets.
Later note, per comments: Let's make $\delta_A$ small enough so that if $x,y\in A$ and $|x-y|<\delta_A$ then $|f(x)-f(y)|<\varepsilon/2$, and if $x,y\in B$ and $|x-y|<\delta_B$ then $|f(x)-f(y)|<\varepsilon/2$.
Let $\delta=\min\{\delta_A,\delta_B\}$.
If $x,y\text{ both}\in A$ or $\text{both}\in B$, and $|x-y|<\delta$ that does it, as above.
If $x\in A$ and $y\in B$, then the distances from $x$ to the boundary point $b$, and from $y$ to $b$, are less than $\delta$, so $$|f(x)-f(y)| \le|f(x)-f(b)|+|f(b)-f(y)|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon.$$
So in all cases, $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon$.