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Let $E$ be a non-empty subset of an ordered space. Suppose that a is a lower bound of $E$ and $\beta$ is a upper bound of $E$. Show that $ \alpha \leq \beta $.

Proof:

(1) If $\alpha$ is a lower bound of $E$ then $\forall x\in E\quad x\geq \alpha$

(2) If $\beta$ is a upper bound of $E$ then $\forall x\in E\quad x\leq \beta$

From (1), (2) and since $E$ is ordered then

$$\alpha\leq x\leq\beta\Rightarrow \alpha\leq\beta$$

I'm starting now with real analysis and I'm still learning the art of demonstrating. Is this right? That's enough?

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    $\begingroup$ Change "since $E$ is ordered, then..." with "since $E$ is non-empty, there is $x\in E$; for such $x$...". And it should then be substantially correct. The claim may not be true for $E=\emptyset$ because all elements of $X$ are vacuously both upper bounds and lower bounds of $\emptyset$. $\endgroup$ – user228113 Apr 3 '17 at 13:00
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It's almost enough. But no, I wouldn't give you full marks for it, because if you are just beginning, it's good to be strict.

And a strict judge is a stupid judge, meaning he sees what you wrote, not what you meant to write. So, a strict judge would look at your proof and ask a simple question:

What is this $x$ that popped up all of a sudden? For which $x$ is the inequality $\alpha\leq x \leq \beta$ true?

Another question I might ask here is (and this is already a hint on how to improve the proof):

Where did you use the fact that $E$ is non-empty? An empty set $E$ also satisfies all your conditions, yet the conclusion does not hold for it, so obviously there's something wrong with your proof!

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  • $\begingroup$ You mean I should mention that $x$ is arbitrary? About your second question, the fact that I used $\forall x\in E$ is not enough to specify that $E$ is non-empty ? What you mean by "Where did you use the fact that $E$ is non-empty?" $\endgroup$ – Roland Apr 3 '17 at 13:11
  • $\begingroup$ @Roland But $x$ can't be arbitrary. For example, if $E=\{0\}$, and $\alpha=-1$ and $\beta= 1$, then the inequality $\alpha\leq x$ is not true for $x=-2$. $\endgroup$ – 5xum Apr 3 '17 at 13:13
  • $\begingroup$ @Roland Saying $\forall x\in E$ does not use the fact that $E$ is non-empty. For example, the statement $\forall x\in\{\}: x<0$ is a true statement. $\endgroup$ – 5xum Apr 3 '17 at 13:14
  • $\begingroup$ I expressed myself wrong, what I meant by arbitrary "$x$" is any $x$ in $E$ $\endgroup$ – Roland Apr 3 '17 at 13:15
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    $\begingroup$ @Roland Yes, that's it! But you have to explain to the stupid judge that the $x$ you are talking about is the $x$ you know exists in $E$! $\endgroup$ – 5xum Apr 3 '17 at 13:19
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The principle you are using is correct. If $E$ is an ordered set then "$\le$" has the transitivity property. Transitive means: $x \le y, y \le z \implies x \le z$ for all $x,y,z \in E$.

In terms of the proof you should always define your variables, so what is $x$?

Here is my example proof: Since $E$ is non-empty, let $x \in E$. Since $\alpha $ is a lower bound for $E$, $\alpha \le x$. Since $\beta$ is an upper bound for $E$, $x \le \beta$. So by transitivity, $\alpha \le \beta$.

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