In Real Analysis class, a professor told us 4 claims: let x be a real number, then:
1) $0\leq x \leq 0$ implies $x = 0$
2) $0 \leq x < 0$ implies $x = 0$
3) $0 < x \leq 0$ implies $x = 0$
4) $0 < x < 0$ implies $x = 0$
Of course, claim #1 comes from the fact that the reals are totally ordered by the $\leq$ relation, and when you think about it from the perspective of the trichotomy property, it makes sense, because claim #1 says: $0 \leq x$, and, $x \leq 0$, and then, the only number which satisfies both propositions, is $x = 0$.
But I am not sure I understand the reason behind claims #2, #3 and #4. Let's analyze claim #2:
It starts saying $0 \leq x$, that means that x is a number greater than or equal to $0$, but then it says $x < 0$, therefore x is less than zero. I think no number can satisfy both propositions, as it would contradict the trichotomy property, because x is less than $0$ AND greater than or equal to $0$.
Same thing with claim #4, since $0 < x < 0$ means: $x > 0$ AND $x < 0$, and no real number satisfy both propositions at the same time. Therefore, saying $x = 0$ is the same as saying $x = 1$, or $2$, or $42$ (this is because the antecedent if always false).
Am I missing something? My professor then told us that these were axioms, but I think that axioms should not contradict well established properties (like the trichotomy property) or, at the very least, make some sense. Are claims #2 to #4 well accepted and used "axioms" in real analysis?
Best Answer
A false proposition implies any other proposition, so given the assumption on the left of 2),3), or 4), which are each false for any real $x,$ one could put any statement after "implies" in these and the overall statement would hold.
Here's a made-up example where a proof could use e.g. claim d) during the proof. The overall goal of the proof would be to show under certain hypotheses that $x=0.$ Suppose somehow the proof split into case A and case B, and that in treating case A one could show each of $x \le 0$ and $0 \le x.$ Here the use of a), namely $0 \le x \le 0$ implies $x=0,$ would suffice and finish case A in a mathematically sound way. On the other hand suppose in case B one could show each of $x<0$ and $0<x,$ thus arriving at the left side $0<x<0$ of claim d). Though it is logically correct to conclude here that again $x=0,$ since $0<x<0$ is false, in my opinion a better mathematical write-up of case B would be, once having arrived at the two statements $x<0$ and $0<x,$ just to say something like "thus case B cannot arise after all" or "so case B is contradictory".
I myself haven't seen a proof by a good mathematical expositor which used anything like claims b), c), or d) during the argument; as outlined in the above made-up proof scenario such proofs would just say such things as "this case does not arise" at the appropriate times.