Does $0 < x < 0$ imply $x =0$?

Question

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?

Answer 1

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.

Answer 2

$A\implies B$ means ($B$ or (not $A$)).

Let $A:0<x<0$ and $B:x=0$.

$A$ is false, therefore $not(A)$ is true. Thus $A\implies B$ is true.

See https://en.wikipedia.org/wiki/Vacuous_truth .

Answer 3

Please bear with me as a layman... but I'll try to explain why I was puzzled (and perhaps also why the OP was puzzled).

The other answers are correct if we assume that the natural language term "implies" used by the teacher as cited by the OP means the logical operation implication, or $\implies$.

That is probably what the teacher, tongue in cheek, indeed meant. What I, at first, understood though was that the first statement defines a set, and the second statement is about that set:

Let $\mathbb{M} = \{x | x \in \mathbb{R}, 0 \lt x \lt 0\}$; then $\mathbb{M} = \{0\}$.

Perhaps the proper notation would be

$\{x | x \in \mathbb{R}, 0 \lt x \lt 0\} = \{0\}$.

That's trivially wrong, hence the confusion.

What the professor meant was probably that the following statements are all true (given that $0\lt x \lt 0$ is always false, we can simply substitute $\bot$):

$\bot \implies x = 0$, or
$\bot \implies x \ne 0$, or
$\bot \implies y=42$,
or anything else.

(I'm not sure whether one can use $\bot$ that way; I may be too influenced by programming.)

Answer 4

You gave the right answer when you mentioned that "the antecedent if always false". You need to come back to the fact that, in logic, $p \rightarrow q$ is true if and only if $q$ is true or the negation of $p$ is true. In (2), (3) and (4), $p$ is false and hence $p \rightarrow q$ is true.

Answer 5

First of all, I disagree these are axioms. What the axioms of set theory, the real numbers, the integers, etc. are can have discrepancies from textbook to textbook, person to person, etc., but axioms should not be taken to mean "a list of obvious properties and I can't be bothered to prove them."

A constructive way to push back against your professor would be to ask him/her for the complete list of axioms they are using. The list should be, well, complete. It shouldn't sprawl. It might include these and omit some other properties I personally think of as axioms.

Next, the first is true and that should make sense by itself.

The next three are vacuously true. They all boil down to "if <some false statement> then <any statement>" which is always true. This is called being vacuously true. $0 < x < 0$ is false for any $x$, so "if $x$ is a real number such that $0 < x < 0, \text{ then }1 = 2$" is a true statement.

Vacuously true statements are considered logically valid but not logically sound.

You can think of it this way: if you ask "for all $x$, if $0 < x < 0$, then $x = 1$ and $x = 7$ at the same time." This is true - you certainly can't find any $x$ satisfying $0 < x < 0$, so there are no counterexamples!

Answer 6

Does 0 < x < 0 imply x = 0? Yes. Does 0 < x < 0 imply x = 1? Also yes. If you don't believe it, prove me wrong. Show me an x such that 0 < x < 0, but x ≠ 1. You can't. There is no such x.

If you have a condition that is false, like 0 < x < 0, then you can draw any conclusion from it.

Answer 7

There's another interpretation for your professor's remarks that doesn't rely on statements (2)-(4) being vacuously true: We can consider $x$ and $0$ as limits of sequences and interpret the inequality symbols as representing pointwise comparisons between the sequences.

For example, the interpretation of (4) would be "If $a_n < b_n < c_n$ and both $a_n$ and $c_n$ have limit $0$ as $n\rightarrow\infty$, then $b_n$ also has limit $0$ as $n\rightarrow\infty$." Taken together, these statements are often called the Squeeze Theorem or Sandwich Theorem, which is an abundantly useful tool in real analysis. It is not an axiom, but rather a consequence of the usual axioms for $\mathbb{R}$.

Granted, this choice of notation would be exceedingly lazy on your professor's part and confusing for you and your classmates. If you haven't discussed limits yet, then this interpretation is implausible. Nonetheless, it's often a valuable exercise to take vacuous claims and try to construe them in a way that gives them substance.