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Consider $(2,3) \subset \mathbb{R}$ and $(3,4) \subset \mathbb{R}$ (They are open intervals).

Would $(2,3) \cup(3,4) = (2,4)$ and $(2,3) \cap(3,4) = \emptyset$?

There is a small section of set theory I need to teach out of an algebra book, and while I understand it, I want to make sure this example I came up with has an appropriate solution. It has been awhile since I have seen set theory, and even though I still remember most of the concepts, there was always an odd example like this one where even though $3$ is not defined, we can still find the union. If my answers are correct, how would you intuitively explain why those are the answers?

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    I'm assuming the question is about open intervals. – 2017-02-06
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    I think he means the open interval $(2,3)$ ? – 2017-02-06
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    Assuming that the notations are for intervals, the first one does not include $3$. – 2017-02-06
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    Sorry they are open interval. – 2017-02-06

2 Answers 2

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$(2,3)\cup(3,4)$ is NOT $(2,4)$ because the element $3$ is not in $(2,3)$ nor $(3,4)$. However, for closed intervals the equality would hold $$[2,3]\cup[3,4]=[2,4]$$ The second statement, i.e. that $(2,3)$ and $(3,4)$ are disjoint is correct.

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Your second part is correct - there are no points in common between $(2, 3)$ and $(3, 4)$, so their intersection is empty.

Your first part, though, is incorrect - the union of $(2, 3)$ and $(3, 4)$ does not contain the point $\{3\}$, because neither of the components includes it. Thus, the result is the interval $(2, 4)$ except for $\{3\}$, or in other words $(2, 3) \cup (3, 4) = (2, 4) \setminus\{3\}$.