counterexample to a proof.

Question

Prove the following statement;

$$\forall a,b \in R (\forall \epsilon > 0 (a \le b + \epsilon) \rightarrow a \le b)$$

I can't see how this is true

This means that I can pick a number for all three variables and this statement should hold, but

Let $\epsilon$ = 100

let b = 0

let a = 1

then $(a \le b + \epsilon)$ holds but $a \le b$ doesn't.

This question is given right after a true/false section, so I'm assuming i'm somehow missing something, and there is a proof for this?? But, right now my reasoning seems sounds, to me.

I think the inner statement reads "if $a \le b + \epsilon$ for all $\epsilon>0$, then $a \le b$," rather than "for any $\epsilon>0$, the statement $a \le b+\epsilon$ implies $a\le b$." — angryavian, Oct 07 '15 at 06:01
It says that if $a\le 10.1$ and $a\le 10.01$ and $a\le 10.000001$, and $\dots$, then $a\le 10$. — André Nicolas, Oct 07 '15 at 06:05
Does this answer your question? Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$? — , Sep 22 '23 at 08:54

score 3 · Answer 1 · answered Oct 07 '15 at 06:05

3

The key is that it needs to be true for ALL $\epsilon >0$. if $\epsilon < 1/2$ for example and b = 0, a = 1 it is not true that $a <= b + \epsilon$.

answered Oct 07 '15 at 06:05

fleablood

130,341

Anders Kaseorg · Accepted Answer · 2015-10-07T06:23:27.793

You may be misinterpreting the statement due to ambiguous parenthesization. It is almost certainly meant to be interpreted

$$\forall a, b \in \mathbb R.\ ((\forall \epsilon > 0.\ a \le b + \epsilon) \rightarrow a \le b).$$

The important thing is that the scope of the quantification of $\epsilon$ should stop before the $\rightarrow$. If you think about the difference between $\forall \epsilon. (p \rightarrow q)$ and $(\forall \epsilon. p) \rightarrow q$, you will see why your reasoning doesn’t apply to the latter.

Why did I infer that the statement should be interpreted this way? Just because this way the statement is true. Notational conventions seem to vary here, and depending on the conventions in use, it’s possible that the statement may be written incorrectly.

score 1 · Answer 3 · answered Oct 07 '15 at 06:02

The statement requires that the inequality holds FOR ANY epsilon greater than 0. So if epsilon is less than $1$ this statement does not hold in the case that you mentioned.

This statement can be proven by contradiction is you assume $a\leq b+\epsilon$ and $a>b$.

score 1 · Answer 4 · answered Oct 07 '15 at 06:18

The example that you considered a counterexample to the proposition is wrong; for, your choice of $a$ and $b$ does not satisfy the hypothesis of the proposition. If $a:=1$ and if $b:=0$, then $a > b$, so $a > b + \varepsilon$ for some $\varepsilon > 0$, say $\varepsilon := \frac{1}{2}$; but the proposition only considers pairs of $a,b$ such that $a \leq b + \varepsilon$ for all $\varepsilon > 0$.

counterexample to a proof.

4 Answers4

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