Can Mathematica Handle Open Intervals? Interval complements?

Question

Open Intervals

Following up on this question, I was wondering whether Mma can handle open intervals. For example, the union of the intervals, $$1<x<5$$ and $$5<x<8$$

should not include the number 5. This is easy enough to do in one's head, but how can it be done, if at all, computationally?

Interval Complement

Also, is there a way to find the complement of two intervals? IntervalComplement[int1,int2,int3] should contain all the points in int1 that are not in the other intervals.

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Edit:

Let's take Mark McClure's data as an example.

int1 = x < -2 || -1 <= x < 1 || x == 3 || 4 < x <= Pi^2;
int2 = -3 <= x < 0 || x > 1;

The intervals are shown below:

The Interval Complement (drawn above in blue on the x-axis) would seem to be:

x < -3 || 0 <= x < 1

Answer 1

I'd represent the sets using inequalities and/or equalities and then apply Reduce. Here's an example:

set1 = x < -2 || -1 <= x < 1 || x == 3 || 4 < x <= Pi^2;
set2 = -3 <= x < 0 || x > 1;
Reduce[set1 && set2]

Here's the complement of the union of the two intervals.

Reduce[!(set1 || set2)]

(* Out: x==1 *)

We might define an interval complement function as follows:

intervalComplement[bigInt_, moreInts__] := 
  Reduce[bigInt && (! (Or @@ {moreInts}))];

For example:

intervalComplement[-10 < x <= 10, -8 < x <= -6, 
  0 <= x <= 2, x == 3]
(* Out: -10 < x <= -8 || -6 < x < 0 || 2 < x < 3 || 3 < x <= 10 *)

Answer 2

This is my code

{a = x > 1 && x < 5, b = x > 5 && x < 8}
{Reduce[a && b]}  

and

{a = x > 1 && x < 5, b = x >= 5 && x < 8}
{Reduce[a || b]}

Edit

Some examples

{a = x > 0 && x < 3, b = x > -1 && x < 2, c = x > -2 && x < 1} {Reduce[a && (b || c)],
Reduce[(a && b) || (a && c)], Reduce[a || (b && c)], Reduce[(a || b) && (a || c)]}