An Abuse of This Theorem with respect to the Supremum of a Set?

Question

There is a theorem stated as follows.

Let $x$,$y \in \mathbb{R}$ and suppose for all $\epsilon > 0$ we have $x\leq y+\epsilon$. Then $x \leq y$.

I was doing some busy work when I noticed a possible abuse of this theorem, and I wish to see what illegal move was made whether explicitly or implicitly.

Let $S$ be a nonempty subset of $\mathbb{R}$ that is bounded from above. By the completeness axiom, the supremum of $S$ exists, call it $x$.

Now let $\epsilon > 0$. Then $x-\epsilon < x$. Because $x$ is the least upper bound of $S$, there exists an $s^{\prime} \in S$ such that

$$x - \epsilon < s^{\prime} < x$$

which implies

$$x - \epsilon < s^{\prime} $$

or

$$x < s^{\prime} + \epsilon \tag{1}$$

But from the above theorem $(1)$ becomes

$$x \leq s^{\prime} \tag{2}$$

which is a contradiction. We cannot even take $x = s^{\prime}$, because this would imply that the supremum of a set is always the maximum, but that is obviously not true.

Is there an invalid move somewhere before reaching $(2)$? I know that because $(2)$ produces a contradiction, we can infer by the Trichotomy Law that $s^{\prime} < x$, but I want to know if there is an illegal move written above or that may be implied from the above reasoning, but it is not explicitly written.

Edit. Made an edit to the theorem to include universal quantification of $\epsilon$ over the positive reals.

Answer 1

Yes, there is a false step. The problem is that your theorem is incorrectly phrased. It should be:

If $\forall\epsilon>0$ we have $x\leq y+\epsilon$, then $x\leq y$.

Notice the $\forall$ quantifier. Note also that $y$ is independent of $\epsilon$ in this theorem.

Now it is true that for any $\epsilon$ you can find an $s'$ such that $x-\epsilon\leq s'$. However this $s'$ depends on $\epsilon$, in fact for any $s'$ that you choose for a given $\epsilon$ such that the inequality holds I can find a smaller $\epsilon$ for which it no longer holds (in the cases where the supremum is not the maximum, if the maximum exists then of course choosing s'=x will work $\forall\epsilon$)

You can therefore not apply the theorem.

Answer 2

You have to understand the theorem very clearly. And for that it is of utmost importance to ditch the greek symbol $\epsilon$. The use of greek symbols somehow tends to make them special (meaning difficult to handle, requiring special treatment so much so that they deserve a tag named "epsilon-delta" on MSE). Sorry these are not special in anyway apart from being positive and are on the same footing as real numbers $a, b, c, x, y, z$. The convention to use greek symbols in such context is well established and one must respect that, but when it comes to understanding them it is best to recast them in mind using usual roman symbols.

So the theorem goes like this:

Let $x, y$ be real numbers with the property that for any/all real number(s) $z > 0$ we have $x \leq y + z$. Then $x \leq y$.

From the statement of the theorem it is obvious that $x, y$ are some fixed numbers (and we are discussing that they possess some specific property which might not be possessed by other pair of real numbers). And the adjective "any" for $z$ means that $z$ can be any positive real number so that the inequality $x \leq y + z$ is actually representing a set of inequalities one for each positive $z$. It is essential that the inequality holds for all positive $z$ and not for some positive value of $z$ and then only we can conclude that $x \leq y$.

Now in your case given any $z > 0$ you are able to pick an $s'$ based on $z$ such that $x < s' + z$. Clearly here $x$ is a fixed number, but both $z,s'$ vary because $z$ can be any positive number and $s'$ depends on $z$. So the theorem does not apply.

---

Moreover note that the theorems of analysis such as the one we are discussing here are trivial and they hold if we change "real" to "rational" in their statement. They are basic facts about the way order relations work in $\mathbb{Q}$ or $\mathbb{R}$. Giving them any more importance than they deserve is only going to create confusion. The only complication in their understanding comes from the use of quantifiers like "any" and the use of greek symbols.

In contrast to this nobody raises an eyebrow when the quantifier "any" is used in the following theorem:

Any natural number greater than $1$ can be expressed as a product of prime numbers in essentially unique way (if we regard order of factors as immaterial).

or

For any natural number $n > 1$ there exists a prime $p$ such that $p$ divides $n$ but no prime greater than $p$ divides $n$.

So I guess the fundamental problem is not really the universal quantifier "any/all" but rather the fact that by habit most people are accustomed to understanding operations $+,-,\times, /,=$ rather than operations like $<, >$.