Can a valid proof by contradiction contradict the opposite proposition?

Question

Sorry if the wording in the title is a bit off, an example will better illustrate my query. Say I wanted to prove the following: $$ (x_n) \text{ and } (y_n) \text{ are both real sequences such that } \forall n \in \mathbb{N}, \ x_n \leq y_n. \text{If } x_n \rightarrow x \text{ and } y_n \rightarrow y, \text{ then } x \leq y.$$

To prove this by contradiction we first assume the opposite proposition $ x > y $ and show that this leads to the statement $ x_n > y_n $, which contradicts one of our initial conditions (namely $ x_n \leq y_n) $.

Could an equally valid proof use the opposite proposition $ x > y $ and show that this leads to $ x = y $, which contradicts only the opposite proposition and not any initial conditions?

Answer 1

I think the short answer is yes: You may have a (very) common misconception about proof by contradiction.

To prove something by contradiction you merely derive any logical contradiction. In basic analysis proofs, it is often clean for your contradiction to be of the form $1 < 0$ or something concrete, as opposed to contradicting the hypotheses.

Answer 2

As you can typically prove anything when working in an inconsistent system, any contradiction, whether it's (most obviously) to some condition in the formulation of the statement, to the assumption made when beginning the proof, to basic arithmetic (i.e $1>2$), ... will be the contradiction you look for (and - in most cases - can be used to "prove" any other contradiction).