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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) ???

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    Uniformly continuous means that given an epsilon > 0, a single delta>0 can be chosen that works simultaneously for all points c in A and B.2012-10-20

2 Answers 2

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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 $|f(x)-f(y)|<\varepsilon$.

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    Pick $\delta_1$, $\delta_2$ as in my answer, except make them small enough to guarantee |f(x)-f(y)|<\varepsilon/2 instead of <\varepsilon. That will take care of all three cases: both points in $A$, both in $B$, and one in $A$ and one in $B$.2012-10-21
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So the formal answer should be as follows?:

Part 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$

Part 2

f is uniformly continuous on A. This means that for every $\epsilon \gt 0$ there exists a $\delta_1 \gt 0$ such that $|x-y|\lt \delta_1$ implies $|f(x)-f(y)|< \epsilon$ for every $x,y$ $\in$ A. Let $\epsilon >0$ be given. Let $\delta_1 $ = $g_1$($\epsilon$), which is small enough to work for the given $\epsilon$

f is uniformly continuous on B. This means that for every $\epsilon \gt 0$ there exists a $\delta_2 \gt 0$ such that $|x-y|\lt \delta_2$ implies $|f(x)-f(y)|< \epsilon$ for every $x,y$ $\in$ B. Let $\epsilon >0$ be given. Let $\delta_2 $ = $g_2$($\epsilon$), which is small enough to work for the given $\epsilon$

Now consider $f$ on the domain of $A \cup B$. Let $\epsilon >0$ be given. Choose $\delta = \min\left \{\delta_1, \delta_2 \right \}$.

Because A and B are connected, and continuous functions preserve connected sets, we can use this $\delta$ for any given $\epsilon>0 $.Namely, for every $\epsilon \gt 0$ there exists a $\delta =\min\left \{\delta_1, \delta_2 \right \} \gt 0$ such that $|x-y|\lt \delta=\min\left \{\delta_1, \delta_2 \right \}$ implies $|f(x)-f(y)|< \epsilon$ for every $x,y$ $\in$ $A\cup B$, concluding that f is uniformly continuous on $A \cup B$

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    I changed \text{sup A} to \sup A. That is standard. It doesn't just prevent italicization of "sup" it also provides proper spacing in things like $5\sup A$, and in a "displayed" setting, it affects positions of subscripts, thus: $\displaystyle\sup_{x\in A}$. And there's no reason for the $A$ to be within \text{}. Similarly I changed min to \min.2012-10-21