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Well... I got a idea from this question If $f$ is uniformly continuous on two sets, show that $f$ is also uniformly continuous on the union of two given sets. So Let me think more the general case. I would suggest you the statement that I made and regard it as true.

$f$ is a uniformly continuous on both $D_1$ and $D_2$ (Here the $D_1 \cap D_2 \not= \phi$) Then, $f$ is a uniformly continuos on $D_1 \cup D_2$

You might look above statement is trivial though, still I can't prove it. :( Does anyone have a idea or hint to prove that?

p.s.) If the statement is false, What is the counterexamples?

han fei
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  • @TomCollinge, your counterexample is incorrect. Intersection of the [0,1) and [1,2] is empty set. But he add condition that intersection is not empty set. – se-hyuck yang Mar 13 '21 at 14:49
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    @se-hyuckyang I missed that thanks. Removed my earlier comment – Tom Collinge Mar 13 '21 at 14:54
  • How about the considering cases like the "[1,2] and [2,3] (intersection is a point)" and "[1,3] and (2,4] "(intersection is a interval) ? Then, you would follow the proof you linked. I considered $D_1$ and $D_2$ are intervals. – se-hyuck yang Mar 13 '21 at 15:15
  • @se-hyuckyang I don't quite follow that, but here's an amendment of my earlier idea in an answer. – Tom Collinge Mar 13 '21 at 15:26

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Let $D_1 = [-1, 0] \cup (1, 2]$ and $D_2 = [0, 1]$
Define $f(x) = 0$ on $[-1, 0]$; $f (x) = x $ on $[0, 1]$; $f(x) = 0 $ on $(1, 2]$
$f$ is separately UC on $D_1, D_2$ but discontinuous on $[-1, 2]$

The key to constructing the counterexample is that $D_1, D_2$ must share some part of their boundary in order to have at least one point in common, but then uniform continuity will be preserved at the intersection. So, it's set up so that $D_1, D_2$ can meet without intersecting somewhere else. One could construct counterexamples in the plane where $D_1, D_2$ are each connected sets and intersect on part of their boundary but only meet on another part. For example let the line $(0, -1)$ to $ (0, +1)$ be the boundary between $D_1, D_2$ with $D_1$ to the left and $D_2$ to the right. Let both include the line $(0, -1) $ to $(0, 0)$ so the intersection is non-empty, but only one include $(0, 0 )$ to $(0, 1)$ with the other being open up to this line.

Tom Collinge
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