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Suppose that $v$ is an upper bound for a set $S$ of real numbers. Prove, by the contrapositive method, that, if there is no real number $\epsilon > 0$ such that, for every element $x \in S$, $x \le v − \epsilon$, then there is no real number $u < v$ such that $u$ is an upper bound for $S$.

Proposition: If there is no real number $\epsilon > 0$ such that, for every element $x \in S$, $x \le v − \epsilon$, then there is no real number $u < v$ such that $u$ is an upper bound for $S$.

A: $v$ is an upper bound for a set $S$ of real numbers. There is no real number $\epsilon > 0$ such that, for every element $x \in S$, $x \le v - \epsilon$.

B: There is no real number $u < v$ such that $u$ is an upper bound for $S$.

My Working

A1 ($\neg B$): There is a real number $u < v$ such that $u$ is an upper bound for $S$.

B1 ($\neg A$): There is a real number $\epsilon > 0$ such that, for every element $x \in S$, $x \le v - \epsilon$.

A2: Let $\epsilon > 0$ be a real number.

A3: $v > u \ge x$ where $x \in S$.

$\implies 0 > u - v \ge x - v$

A4: $\epsilon > 0 > u - v \ge x - v$

$\implies 0 > -\epsilon > u - v - \epsilon \ge x - v - \epsilon$

$\implies v > v - \epsilon > u - \epsilon \ge x - \epsilon > x$

$\implies v - \epsilon \ge x$

$Q.E.D.$

I would greatly appreciate it if people could please take the time to review my proof for correctness. If there are any errors, then please specify what the error is, why, and a correction.

1 Answers 1

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Wrong. It's false that $x-\epsilon>x$ since $\epsilon>0$. Actually your proof is wrong from the beginning. You have to prove that $\exists \epsilon>0$ so giving some arbitrary $\epsilon$ isn't a good start. The proof is next:

Suppose $v>u\geq x,\forall x\in S$ (This is $v>u$ and $u$ is upper bound of $S$) and define $\epsilon=v-u$. Since $v>u$, then $v-u>0$ so $\epsilon>0$ by definition. Also we have from the hypothesis that $v-\epsilon=u\geq x,\forall x\in S$.

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    Thanks for the reasons. You're right: $x - \epsilon \not > x$. So my primary mistake was to let $\epsilon > 0$? Also, can you please elaborate on your solution? How did you get $\epsilon = v - u$? I don't see anywhere that says $\epsilon =$ something? – 2017-02-17
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    This is the point of proof. When you have to prove that there is some #%@, that means you need to explicit find an expression such that satisfices the properties you want. So in this case I found that taking $\epsilon=v-u$, it satisfices both conditions I want – 2017-02-17
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    But I can say, "Let $\epsilon$ be a real number", right? I just can't make it be $> 0$? – 2017-02-17
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    No, this is the mistake. What you need is to develop the hypothesis $(A_1)$ and then construct $\epsilon$ from this. You don't set $\epsilon$ until you know which value satisfices the properties of the conclusion you want to prove, so you can't let $\epsilon$ until the end – 2017-02-17
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    Your answer makes sense. But how did you know to define $\epsilon = v - u$? – 2017-02-17
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    Proofs of existence demands purely intuition since you can't just play with the hypothesis like contradiction for example. To develop your intuition in this problem, as we say (my classmates and me) you have to "master it" ("maestrearlo" in Spanish). That is first think a fast proof in your mind (or to say it) avoiding formality, just to understand the problem. A kind of sketch of proof. Also, if you can, you might try to draw the problem and try to find some "geometric" solution, especially when you are working with sets. – 2017-02-17