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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 , $\text {inf B} \le x \le \text {sup A}$ implies $x = \text{sup A}$. I chose x arbitrary, so $ A\cap B = \text {sup A}$

(2)???

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Start by writing down what "uniformly continuous" means. –  Chris Eagle Oct 20 '12 at 23:03
    
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. –  MSKfdaswplwq Oct 20 '12 at 23:11
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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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@MichealHardy: Dont we have to talk about the connectedness of A and B? –  MSKfdaswplwq Oct 21 '12 at 11:36
    
Suppose we let $\delta=\min\{\delta_1,\delta_2\}$. Maybe the remaining issue is this: what if $|x-y|<\delta$ and $x\in A$ and $y\in B$. The argument I gave above works if $x,y\in A$ or $x,y\in B$. Let $b$ be that boundary point between $A$ and $B$, which, under your hypotheses, is a limit point of both sets. Then $x$ differs from $b$ by less than $\delta$ and $y$ differs from $b$ by less than $\delta$, so each of $f(x)$, $f(y)$ differs from $b$ by less than $\varepsilon$, so $|f(x)-f(y)|<2\varepsilon$. Not good enough, so here's what I propose to do: [out of space; continued below] –  Michael Hardy Oct 21 '12 at 21:55
    
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 $B$. –  Michael Hardy Oct 21 '12 at 21:57
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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. –  Michael Hardy Oct 21 '12 at 17:07
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