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Answer by Naba Kumar Bhattacharya for Prove that $|(a,b) \cup (c,d)| = (b-a)...
If $(a,b)\cap (c,d) \neq \emptyset$ then, by considering cases one can see that the outer measure of $(a,b)\cup (c,d)$ is not the same as $b-a + d-c$.For the other side, assume they are disjoint,...
View ArticleAnswer by Martin R for Prove that $|(a,b) \cup (c,d)| = (b-a) + (d-c) \iff...
We have$$\begin{align}& \, (b-a) + (d-c) - |(a,b) \cup (c,d)| \\ =& \, |(a, b)| + |(c, d)| - |(a,b) \cup (c,d)| \\ =& \, |(a,b) \cap (c,d)|\end{align}$$where the last identity is an...
View ArticleProve that $|(a,b) \cup (c,d)| = (b-a) + (d-c) \iff (a,b) \cap (c,d) =...
This is from Section 2A, exercise 7 in Axler's MIRA book.Suppose $a,b,c,d$ are real numbers with $a<b$ and $c<d$. Prove that$$|(a,b) \cup (c,d)| = (b-a) + (d-c) \iff (a,b) \cap (c,d) =...
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