>>7619806Ok, I'm only going to do one direction for open intervals, but generalizing and proving the converse will be just as easy.
so this is a proof of a biconditional. We have two statements A and B, and we want to prove that A is true if and only if B is true. In order to do this we have to prove first
"A then B"
and then we prove
"B then A"
Ok, so the forward direction. I want to prove "if I is an interval, then for any x and z in I and y a real number satisfying x<y<z, y is in I" There are going to be three ways to do this. The first is to prove it directly, that is to assume that "I is an interval" and prove from that if x<y<z, then y is in I. A second way is to prove the contrapositive, that is assume "I is not an interval" and show that there exists a triplet (x,y,z) such that x<y<z and y is not in I. The third way is by contradiction, that is to assume both "I is an interval" and "there exists a triplet (x,y,z) such that x<y<z and y is not in I."
Because I have intuition for this type of proof, I know that you can prove it easily directly:
"Let I=(a,b) be an interval and x<z two elements in it. Suppose x<y<z."
This is my assumption for the proof.
"Then we know from the definition of an interval that a<x<y<z<b,"
Here I'm using the definition of an open interval, combined with the fact that x<y<z.
"so a<y<b, and since I contains all reals r satisfying a<r<b, we know y is in I."
This is basically a fact that comes for free once we know the statement a<x<y<z<b.
Can you prove the converse? Here's my hint: begin by assuming "I is not an interval."
