Show that the intersection of any two intervals is an interval

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.