Real Analysis Intervals Proof

[jacobsh47]

The problem is:
Prove that I<J if and only if x<y for every x in I and y in J.

The text says: If I=[r,s] and J=[u,v], I<J means that s<u.

Here is my proof:

Proof by contradiction: Assume that I<J if x>y for every x in I and y in J.
Let I be the interval [r,s] and J be the interval [u,v].
x in I means r <= x <= s
y in J means u <= y <= v
Since x>y, u <= y < x <= s
Therefore u<s
However the definiton of I<J is s<u
This contradicts our assumption
Therefore the original statement is true, I<J if and only x<y for every x in I and y in J.

I just wanted to make sure that this proof is correct and complete.

 

[daon]

The problem is:
Prove that I<J if and only if x<y for every x in I and y in J.

The text says: If I=[r,s] and J=[u,v], I<J means that s<u.

Here is my proof:

Proof by contradiction: Assume that I<J if x>y for every x in I and y in J.
Let I be the interval [r,s] and J be the interval [u,v].
x in I means r <= x <= s
y in J means u <= y <= v
Since x>y, u <= y < x <= s
Therefore u<s
However the definiton of I<J is s<u
This contradicts our assumption
Therefore the original statement is true, I<J if and only x<y for every x in I and y in J.

I just wanted to make sure that this proof is correct and complete.

You need to be VERY careful when negating statements. The negation would be "x>=y for some x in I, some y in J."

This is also an IF AND ONLY IF, which means you need to prove the forward implication and its converse.