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Prove that this is no real number such that $x \leq a$ for all real $x$.

I want to know if the way I proved it is valid or not.

Proof. We first prove that there is no real number $a$ such that $x=a$ for all real $x$. If this were true, then all real numbers would be equal to $a$. This is not possible, because the axiom that garantees the existence of identity elements tells us that the set of real numbers has the numbers $0$ and $1$. Therefore, the set of real numbers has more than only one element.

If $x0$. It can be proven that $1>0$. There is an axiom that tells that if two real numbers are in $R^{+}$, the sum of the two is in $R^{+}$. So, $a+1>0$ is in in $R^{+}$ and in $R$. But we assumed that $a$ is greater than all the real numbers. Thus, $a>a+1$, which is an absurd.

Is this proof correct? Can I improve it? Thank you.

  • 4
    Why not just say that if such $a$ exist then since $a+1$ is also a real number (by axiom, since both $a$ and $1$ are real) and if $a+1\leq a$ it would imply $1\leq 0$ which you said you can prove otherwise. (the first part is not relavent as far as I can tell)2012-08-08
  • 8
    Your proof is no good at all. "$x\le a$ for all $x$" means "$(x < a \hbox{ or } x = a)$ for all $x$". This is not the same as "$(x$x)$ or $(x=a$ for all $x)$", as you seem to think. – 2012-08-08
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    @MJD You are of course correct, but this makes little change to the proof. OP needs to change "more than only one element" to "at least one element different from a", and "for all real x" to "for all real x different from a".2014-07-07
  • 0
    I think you are missing a step. Why is a>a+1 absurd? This should be proven since it's the heart of your proof. How to prove it depends on what axioms you can use.2014-07-07

3 Answers 3

12

Let me propose an alternative proof.

Proof: By contradiction, suppose such an $a \in \mathbb{R}$ exists. Then $a + 567 \le a$, which implies $567 \le 0$, a contradiction.

  • 0
    I do not know what you mean; how would it follow that $a + 567 \leq a$? (More to the point, this fact *depends* on the property of the reals which I described. Since you didn't use it, your conclusion could not work.)2012-08-08
  • 0
    I'm supposing by contradiction that there exists a real number $a$ such that $x \le a$ for all $x \in \mathbb{R}$.2012-08-08
  • 1
    I think @user1296727's point is that it's not a priori clear that $a+567\leq a$ is a contradition until you know some facts about $\leq$.2012-08-08
  • 0
    In detail: the real numbers, with an element $a$ adjoined to them with the property you ascribed, do not form a field. In particular, the proposition that $a \leq b \implies a + c \leq a + c$ is not (immediately) valid; you implicitly relied on this fact in your answer.2012-08-08
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    @user1296727, if $a < b$, then $a + c < b + c$ by the additive property, and if $a = b$, then $a + c = b + c$ since $+$ is a function.2012-08-08
  • 0
    You have not proven (and couldn't prove!) that the additive property applies to your infinite element. This is my complaint; notice that in my proof, I *never* assumed so much about this element. When you "extend" the reals, you are tossing in some guy named "$a$", whose ordering properties have not been explored or proven. Indeed, a system with $a$ in it is *still* consistent, and can have the supremum property (check!); the problem is that this guy "$a$" doesn't have the additive property. (He wasn't part of our operation back when we constructed the reals to have the additive property!)2012-08-08
  • 0
    (My last comment was directed @Brian, as is this one.) In summary, *things only have the properties which you prove about them*. $a$ was never proven to be additive (in the sense that $b < c \implies a + b < a < c$, or vice-versa); although we can meaningfully define the arithmetic operations with respect to it, we *cannot* simply ascribe properties to it which we did not show that it has. This was your fallacy; it is a natural fallacy, but a fallacy nonetheless.2012-08-08
  • 1
    Not quite getting you. I have said that $a$ is a real number, and the additive property is an order axiom. See:http://www.calvin.edu/~rpruim/courses/m361/F03/overheads/real-axioms-print-pp4.pdf2012-08-08
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    @Brian There *are* [non-Archimedean fields](http://en.wikipedia.org/wiki/Non-archimedean_field); your proof, if valid, would contradict this, since it uses only the field axioms to prove inconsistency (i.e., you "prove" that all fields, which have the additive property, must be Archimedean). Maybe you haven't done this already (which would explain the confusion), but we *construct* the reals to study them; when we construct them, we prove that they have certain properties. But I already explained this. I showed that the field axioms, *coupled* with the supremum property, force Archimedean.2012-08-08
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    Without both of these in force, we cannot prove that a field is Archimedean. (Please *do* read my answer; I'm not trying to be a jerk, I'm just trying to clarify.) Also, point of order: in the above, I meant that I assumed the field axioms *for the ordinary reals* (i.e., not "$a$").2012-08-08
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    What do you even mean by "ordinary reals"? I have repeatedly said that $a$ is a real number. And yes, for any $a, b, c \in \mathbb{R}$, $a \le b \implies a + c \le b + c$ is a valid consequnce of the field and order axioms of the real numbers.2012-08-08
  • 0
    Why don't you tell me what you think $a$ is?2012-08-08
  • 0
    $a$ is, of course, precisely as the OP characterized it. It is some hypothetical element of the "reals" larger than any other. If we *assume* that the field axioms are satisfied for all "reals", including *a*, then what we thought were the "reals" cannot be $\mathbb{R}$; this is because we would then have a non-Archimedean field. I proved that $\mathbb{R}$ is an Archimedean field. Therefore, you cannot assume that the field properties (like $b < c \implies a + b < a + c$) hold for this $a$, since its existence makes your field non-Archimedean by default. Your proof therefore does not work.2012-08-08
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    In short: nothing more need (or will) be said about your proof than that it cannot possibly apply to $\mathbb{R}$.2012-08-08
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    As long as $a \in \mathbb{R}$, all the field axioms and order axioms hold for $a$. The reason $a$ cannot exist is why I got a contradiction. And what you wrote isn't even a field axiom.2012-08-08
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    The OP made it clear what $a$ is. No need to use quotes.2012-08-08
  • 6
    The desired property has nothing to do with the field being Archimedean and everything to do with the field being ordered, seeing as it holds for any ordered field, not just the Archimedean ones.2012-08-08
  • 2
    Googling for ["ordered field" smallest element](http://www.google.com/#&q=%22ordered+field%22+smallest+element&oq=%22ordered+field%22+smallest+element) gives, for example, proof in Elements of Real Analysis By Charles Denlinger [p.14](http://books.google.com/books?id=5YhXVE9slLsC&pg=PA14) of the fact that an ordered field does not have smallest/largest element; the proof goes by shifting the element by one; i.e., similarly as in this answer.2012-08-08
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    @MartinSleziak I simply do not take your meaning. There *are* ordered fields with precisely that property. They have the very unambiguous name of [non-Archimedean ordered fields](http://en.wikipedia.org/wiki/Non-Archimedean_ordered_field). My criticisms here are correct (I did pause on them!). This answer starts by assuming that the number is part of an ordered field (clearly, since he assumes that the field axioms apply, and their associated ordering properties). He concludes that this is contradictory, because the ordering does not resemble that of the reals. Of course, it couldn't! ...2012-08-08
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    @MartinSleziak You simply could not prove that an ordered fields are Archimedean on the basis of the field axioms. So, is there something that I am missing in your comment?2012-08-08
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    @user1296727 what you are missing is that this question has nothing to do with Archimedean field at all.2012-08-08
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    @user1296727: The fact that there is no greatest element in the field does not imply that the field is Archimedean. In fact, every ordered field has no greatest element, no matter whether it is Archimedean or not.2012-08-08
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    BTW the implication $a\le b$ $\Rightarrow$ $a+c\le b+c$ is part of the definition of [ordered field](http://en.wikipedia.org/wiki/Ordered_field). (This was the part of the proof which you objected to.)2012-08-08
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This property is a consequence of the construction of $\mathbb{R}$. In particular, it follows from the fact that $\mathbb{R}$ has no upper bound: indeed, the existence of your $a$ would imply that $\sup \mathbb{R} \in \mathbb{R}$. But this is false, since $\mathbb{N} \subset \mathbb{R}$ and $\mathbb{N}$ has no upper bound, because $n+1 > n$ for any $n \in \mathbb{N}$.

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    The closed interval $[0,1]$ has an upper bound although $[0,1)\subset[0,1]$ and $[0,1)$ has no upper bound, because $\frac{x+1}2\gt x$ for every $x$ in $[0,1)$.2012-08-08
  • 1
    I don't understand your comment, sorry. If $x \leq a$ for every $x \in \mathbb{R}$, then $\sup \mathbb{R} \leq a$. Moreover, please remark that $1$ is an upper bound for $[0,1)$. It seems you confuse upper bound, least upper bound, and *maximum*.2012-08-08
  • 1
    Surely you noted that my comment follows exactly your answer: replace $\mathbb R$ by $[0,1]$ and $\mathbb N$ by $[0,1)$ and everything you state about $\mathbb R$ and $\mathbb N$ holds about $[0,1]$ and $[0,1)$. Ergo, your *proof* is bogus. More constructively, you might want to wonder what makes you accept that $[0,1)\subset[0,1]$ has an upper bound, but not $\mathbb N\subset\mathbb R$.2012-08-08
  • 1
    Well, if $x \in [0,1)$, then $x+1 \notin [0,1)$, and this is rather important, isnt'it? My "proof" used the fact that $n+1 \in \mathbb{N}$ whenever $n \in \mathbb{N}$. Now, if $a$ is as in the statement, by definition of $\mathbb{R}$ there exists $\bar{n} \in \mathbb{N}$ such that $\bar{n} \leq a < \bar{n}+1$, and this is a contradiction. However, in my opinion, it is hard to write a proof if we do not agree on what we know about $\mathbb{R}$.2012-08-08
  • 0
    *Well, if $x \in [0,1)$, then $x+1 \notin [0,1)$, and this is rather important, isnt'it?* This is completely off-topic, as already explained at the end of my first comment. Anyway, since you are not listening, I stop here.2012-08-08
  • 1
    I still do not understand. My comment is not off-topic at all! Who cares if $\frac{x+1}{2}>x$ for $x \in [0,1)$? The point is that $\mathbb{N}$ is closed under the addition of a *fixed* positive quantity, viz. $1$. Essentially, my "proof" boils down to Brian's proof: just remark that $a+1 \leq a$ implies $1 \leq 0$. Again: why do you say that $[0,1)$ has no upper bound?2012-08-08
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    @did $[0,1)$ has the upper bound $1$. (You are the one who is not willing to listen here).2012-08-08
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    @did (do you have $\max$ in mind when Siminore said 'upper bound'? This is not the same)2012-08-08
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    @did _and_, come to think of it, I do have to admit I'm quite surprised and extremely nonplussed that two of your inappropriate comments even got an upvote. This looks like fan post and actually worries me a lot.2012-08-08
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    @Thomas Too bad.2012-08-08
  • 0
    Let me say that I did not intend to give *the most economic* proof of the world. My suggestion was: 1) remember that $\sup \mathbb{N} = +\infty$ and 2) remark that $\mathbb{N} \subset \mathbb{R}$. I think that 1) is known much time before $\mathbb{R}$ is constructed. However, the proof that $\mathbb{N}$ contains numbers of arbitrarily large size in some sense proves also that fact that $\mathbb{R}$ is unbounded from above. By the way, some textbooks prove that $\sup \mathbb{N} = +\infty$ by using the archimedean property of $\mathbb{R}$, as in Giaquinta and Modica's book.2012-08-09
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    @Siminore: Economy is not the point, pure logic is. For at least the fourth time (and after a refreshing, but somewhat distracting, interlude by Thomas), let me repeat: that $E\subset F$ and that $E$ has no upper bound in $E$ does not imply that $F$ has no upper bound in $F$. Sure, if $E=\mathbb N$ and $F=\mathbb R$, then $E\subset F$ and $E$ has no upper bound in $E$ and $F$ has no upper bound in $F$. But if $E=\mathbb N$ and $F=\mathbb R\cup\{+\infty\}$, $E\subset F$ and $E$ has no upper bound in $E$ and $F$ has an upper bound in $F$. Which shows your argument alone cannot work on its own.2012-08-09
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    Well, I thought that the whole discussion was about $\mathbb{R}$. Please point out where I try to extend my reasoning to more general sets than $\mathbb{R}$. I can read no sentence about general subspaces or other ordered sets, in my answer.2012-08-09
  • 0
    Of course the whole discussion is about $\mathbb R$! (And *please point out where* I say that *you try to extend your reasoning to more general sets*...) But there is a principle in mathematics that if a *proof* only uses facts that are valid in a given setting and if the result obtained is false in at least one case belonging to this setting, then the so-called *proof* is not valid. **Every** argument in your answer applies equally well to the $(E,F)$-setting I described (if you think that some do not, please show them), ergo... // (And please use the @ thing if you want me to get pinged.)2012-08-09
  • 0
    @did No, I do not think so. $\mathbb{N}$ is unbounded from above, while $[0,1)$ is *not*. This discussion seems incredible to me. I just remarked the fact, which should be known *before* introducing real numbers, that a subset of $\mathbb{N}$ is bounded from above if and only if it contains a largest element. This is definitely *false* for subsets of $\mathbb{R}$ like $[0,1)$.2012-08-09
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    It is incredible, as you say, because you do not listen. OK, enough with me.2012-08-09
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    I listen, but I think you do not want to remark that I used a *specific property* of $\mathbb{N}$: a subset is bounded from above if and only if it has a largest element. My "proof" is heavily based on this, and it cannot be generalized to arbitrary subsets of $\mathbb{R}$, as you suggest. The property I use is equivalent to the induction principle. I did *not* say that "there exists a subset of $\mathbb{R}$ without upper bound". I definitely used $\mathbb{N}$, and its most important order property.2012-08-10
-3

This is a consequence of the Archimedean property of the reals. (I assume that you have some familiarity with real analysis; in particular, I assume that you know that the reals have the supremum property.)


Theorem: For every real $\alpha > 0$ and real $\beta$, there exists an integer $n$ (in the naive sense of integers) such that

$$n \alpha > \beta$$

Proof: Suppose that there were to exist counterexamples $\alpha_0$ and $\beta_0$ (i.e., $\alpha_0$ such that for every integer $n$, $n\alpha_0 < \beta_0$). If we form $S = \{s \in \mathbb{R}: s = n\alpha \text{ for } n \in \mathbb{N}\}$, then this would say that every member of S is smaller than $\beta_0$ by hypothesis.

But clearly $S$ is not empty if $n\alpha_0 < \beta_0$ for every $n \in \mathbb{N}$; since it is bounded above (and since these are real numbers!) there exists $\lambda = \mathrm{sup}(S)$.

Now, for every $\epsilon > 0$, we must have some $s' \in S$ such that $s' > \lambda - \epsilon$. (Otherwise, every $s \in S$ is smaller than $\lambda - \epsilon$, so $\lambda - \epsilon$ would be an upper bound of $S$ strictly smaller than $\lambda$, which was by definition the smallest upper bound. This would be absurd.)

But this means that for some $s' \in S$, $s' > \lambda - \alpha$ (since we put $\alpha > 0$). Since $s' \in S$ if, and only if, $s'$ is an integer multiple of $\alpha$, this says

$$ s' = n\alpha > \lambda - \alpha $$

But then we get

$$ (n+1)\alpha > \lambda $$

so that some member of $S$ is greater than $\lambda$. This is absurd; so we conclude that there are no counterexamples to the theorem in $\mathbb{R}$.

Q.E.D.


Now, the result you want is an immediate corollary.


Theorem: There is no $\gamma \in \mathbb{R}$ such that $\delta < \gamma$ for all $\delta \in \mathbb{R}$.

Proof: Suppose there were a counterexample $\gamma_0$. Pick any real $\delta_0$. By the Archimedean property,

$$\exists n: n\delta_0 > \gamma_0.$$

This directly contradicts the hypothesis, since $n\delta_0$ is a real greater than $\gamma_0$.

Q.E.D.

  • 5
    Why do it this needlessly complicated way, instead of just using that $\mathbb{R}$ is an ordered field?2012-08-08
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    @Tobias Nothing about $\mathbb{R}$ being an ordered field necessarily guarantees the absence of infinite, or infinitesimal, elements. There are [non-Archimedean fields](http://en.wikipedia.org/wiki/Non-Archimedean_ordered_field) in which infinite elements do indeed exist. You cannot derive a contradiction simply from having a totally ordered field.2012-08-08
  • 0
    @Tobias And to be quite frank, since I've already had this discussion at length, and would like to *stop* having it, I'm not going to rehash it with you. If you find my explanation insufficient, see the comment thread with the first user below.2012-08-08
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    Your comments there are not actually correct. In any ordered field $F$, $a < b$ and $c < d$ implies that $a+c < b+d$, so even though the field can contain elements greater than any real, it cannot contain an element larger than all others, which is the issue here.2012-08-08
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    @user1296727: This has nothing to do with whether an ordered field is Archimedian or not. Take "the" hyperreals for example. There is no hyperreal larger than every hyperreal, and the hyperreals are non-Archimedian and contain infinitesimals and infinite elements.2012-08-08
  • 2
    @user1296727 The question is not about the existence of *infinite* elements but, rather, the existence of a *maximal* element. Ordered fields can have infinite elements, but they cannot have a maximal element, since $\rm\:x+1 > x\:$ for all $\rm\,x.\ \ $2012-08-08