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So i've come across this question, with a follow up question of showing that the union of any two intervals need not be an interval.

I don't see how this could possibly be the case. The general structure of my proof would be to consider the case where:

  1. The intersection yields an empty set;
  2. The intersection yields a set with 1 element;
  3. The intersection yields a set with 2 or more elements;

and then consider each case and show that each is an interval.

But a union can only yield one of those three possibilities too. To elaborate: If any of the three possibilities were not an interval, then an intersection is not necessarily an interval, so each of them must be an interval. But each of these possibilities for the intersection are also the only possibilities for a union, meaning a union of intervals must be an interval too, which is not true.

I'm obviously wrong, but why?

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    $\begingroup$ $(0, 1) \cup (1,2)$ is not an interval $\endgroup$ – steven gregory Mar 30 '16 at 6:05
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    $\begingroup$ 1) that does not look anything like an interval to me. That is only three separate points. An interval is all the values between two points. That is three points with none of the values between them. 2) that isn't even close to being the union of (0,1) and (1,2). The union is the set of all points in either of them. That would be all the points between 0 and 1 and all the points between 1 and 2. That would be all the points between 0 and 2 except 0, 1, and 2. Weirdly, you listed exactly all the points that are not in the union and none that are. $\endgroup$ – fleablood Mar 30 '16 at 7:01
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    $\begingroup$ You are having a semantic issue "consider the case" does not mean the case requires the result. Suppose it said prove p is prime. Consider the two cases: p less than or equal to 2, and p >2. now prove q is not prime by considering the same two cases. The case don't mean p or q is or isn't prime. The cases just mean to consider them. $\endgroup$ – fleablood Mar 30 '16 at 7:19
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    $\begingroup$ 3) isn't nescessarily an interval. {0,1,2} isn't an interval. But {0,1,2} is never the intersection of two intervals. $\endgroup$ – fleablood Mar 30 '16 at 7:22
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    $\begingroup$ BTW it depends on your course whether to consider case 1 or 2 intervals. If not the statement isn't true as case 1 and 2 are always possible. Also by the way, if 1 or two are intervals then case 1 or 2 are NOT possible with unions. $\endgroup$ – fleablood Mar 30 '16 at 7:25
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By definition, a set $A\subset{\mathbb R}$ is an interval if $$\forall x, \ y\in A,\quad\forall t\in{\mathbb R}:\qquad x\leq t\leq y\quad\Rightarrow\quad t\in A\ .\tag{1}$$ It is then obvious (on logical grounds, no case distinctions needed) that the intersection of two intervals is an interval. Of course it is allowed to go through the motions anyway:

Let $A$ and $B$ be intervals, let $x$, $y\in A\cap B$, and assume $x\leq t\leq y$. Then $t\in A$ as well as $t\in B$, hence $t\in A\cap B$. This shows that $A\cap B$ passes the test $(1)$.

Note that the claim would not be true if we would not accept the empty set as an interval.

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    $\begingroup$ Thanks for this answer, Christian. I'm very new to proof writing, is it acceptable in proofs to state it is obvious from the definition, or is there further work you would have to do? Further, was my plan of looking at case distinctions okay, or is that considered sloppy work? $\endgroup$ – daviegravee Mar 30 '16 at 11:13
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In the first case, where you get an empty intersection, you do not get an interval. Consider, as in @StevenGregory's example, $(0, 1) \cup (1, 2)$. The intersection is the empty set. How do you represent this as an interval? $(1, 1)$? $(0, 0)$? $(x, x)$ for any real $x$? How would you distinguish between these? So when you say, "an intersection is not necessarily an interval", you are right.

Perhaps you are confusing the notion of sets of real numbers with intervals: all intervals are sets, but all sets aren't intervals. The intersection and union of any two sets is always a set (even though it might be empty), but this does not hold true for intervals.

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  • $\begingroup$ These links seem to suggest that the empty set is an interval: 1. Stack thread 2. Wolfram $\endgroup$ – daviegravee Mar 30 '16 at 6:24
  • $\begingroup$ @d_stewart Yes, but note that in the first one, it says "the definition is satisfied vacuously and $\varnothing$ is an interval." Vacuously satisfied definitions do not always make sense, e.g. an empty set satisfies any condition/property of its elements. $\endgroup$ – shardulc says Reinstate Monica Mar 30 '16 at 6:29
  • $\begingroup$ Thanks @shardulc, the terminology gets lost on me. $\endgroup$ – daviegravee Mar 30 '16 at 11:14
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The problem is: is true that the union and intersection satisfies one of three options, but not necessarily the same (and not necessarily are the same set). For example, in $\mathbb{R}$, the intervals $(0,1), (4,5)$ have empty intersection (and is an interval) and the union is clear that isn't an interval. In $\mathbb {R}^n$, consider the cubes $Q_1=[0,1]^n, Q_2=[1/2,3/2]^n$. The intersection is the cube $[1/2,1]^n$ (an interval) and the union is not an interval.

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Those cases are not the answers but only the cases to consider while finding the answer.

For intersection:

1) if intersection has no points, show the empty set is an interval.

2) if intersection has 1 point, show that a set with 1 point is an interval.

3) if the intersection has at least two points, show that the intersection actually has several me points and those points make an interval.

Now with union

1) the union of two intervals has zero points. This can only happen if the two intervals are both the empty set, but if so, this is an interval.

2) the union has exactly one point. This only happens if the two intervals are both the same single point. If so, this is an interval.

3) the union has two or more points. If so nothing can be determined. It's possible this is an interval, but it's also possible this is two separate disjoint intervals.

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