Union and Intersection of intervals

Question

I have two sets $A = \{1 \leq x \leq 5\}$ and $B = \{5 \leq x \leq 8\}$. Now I want to find the Union and Intersection of $A$ and $B$.

I tried Union[A, B], I got {1 <= x <= 5, 5 <= x <= 8} and for Intersection[A, B], I got {}. The correct answer for $A \cup B$ is [1, 8] and $A \cap B$ is {5}. How do I tell Mathematica to do that?

And if $A = \{1 < x < 5\}$ and $B = \{x > 5\}$. Now I want to find the Union and Intersection of $A$ and $B$. How do I tell Mathematica to do that?

You will want to use Interval. – DavidC Oct 01 '12 at 01:13 — DavidC, Oct 01 '12 at 01:13

score 10 · Accepted Answer · answered Oct 01 '12 at 01:19

10

a = Interval[{1, 5}];
b = Interval[{5, 8}];
IntervalUnion[a, b]

Interval[{1, 8}]

IntervalIntersection[a, b]

Interval[{5, 5}]

answered Oct 01 '12 at 01:19

DavidC

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Interval[{1,5},{5,8}] also gives IntervalUnion[a,b] (+1) – kglr Oct 01 '12 at 01:28
Nice. I wasn't aware of that. – DavidC Oct 01 '12 at 01:30
If A = {1 < x < 5} and B = {x > 1}. How must I do? – minthao_2011 Oct 01 '12 at 01:45
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@minthao: B = Interval[{1, ∞}]. – J. M.'s missing motivation Oct 01 '12 at 01:52
If A = {1 < x < 5} and B= {x > 5}, then Intersection A and B is the empty set,
A = Interval[{1, 5}]; B = Interval[{5, [Infinity]}]; IntervalIntersection[A, B] Mathematica out put Interval[{5, 5}]
– minthao_2011 Oct 01 '12 at 02:07
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@minthao. Mathematica seems to work only with closed intervals. Interval[{1,5] includes the numbers 1 and 5, and therefore does not match your original inequality, 1<x<5. I don't know whether there is a way to not include the end points. Perhaps this is a question worth asking. – DavidC Oct 01 '12 at 02:31
@DavidCarraher I believe that the design objectives for Interval are focused to ease the implementation of interval arithmetic (an area of numerical analysis). See, for example, http://reference.wolfram.com/mathematica/guide/IntervalArithmetic.html – Mark McClure Oct 01 '12 at 02:58
@MarkMcClure Yes, it seems well equipped for dealing with issues of precision. I was thinking more of the logic of intervals in pure mathematics, I suppose. – DavidC Oct 01 '12 at 03:00
My idea is solve Solve[{(x - 1)*(x - 5) < 0, x - 1 > 0}, {x}] – minthao_2011 Oct 01 '12 at 03:26
@minthao_2011 For some reason that code does not work. – DavidC Oct 01 '12 at 03:38
Can I post a new question about this problem? – minthao_2011 Oct 01 '12 at 03:40
Go right ahead. Did you notice that I already posted a follow-up question? – DavidC Oct 01 '12 at 03:47
let us continue this discussion in chat – minthao_2011 Oct 01 '12 at 03:53

mikea · Answer 2 · 2020-04-28T19:07:23.450

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Open/closed intervals compatible with regions:

OpenInterval[a_, b_] := ImplicitRegion[a < x && x < b, {x}];
ClosedInterval[a_, b_] := ImplicitRegion[a <= x && x <= b, {x}];

Semi-open intervals and infinity intervals are similar.

Need some typing to transform region to an expression:

open intervals:

RegionMember[RegionUnion[OpenInterval[0, 5], OpenInterval[5, 8]], {x}] // FullSimplify

gives:

x \[Element] Reals && (0 < x < 5 || 5 < x < 8)

closed intervals:

RegionMember[RegionUnion[ClosedInterval[0, 5], ClosedInterval[5, 8]], {x}] // FullSimplify

gives:

0 <= x <= 8

edited Apr 28 '20 at 19:07

answered Apr 28 '20 at 16:30

mikea

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+1. That approach works in higher dimensions. – user64494 Apr 28 '20 at 18:43
And also works for Integers instead of Reals. – István Zachar Jan 12 '22 at 20:15

Union and Intersection of intervals

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