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I was browsing a proof here. At the end of the proof, the author says:

$\operatorname{diam}(\bar{E}) \leq \operatorname{diam}(E)+ε.$ Since $\epsilon > 0$ is arbitrary, $\operatorname{diam}(\bar{E}) \leq \operatorname{diam}(E)$.

I'm not sure how the author made $\epsilon$ go away by saying that it's arbitrary. Even if it's arbitrary, it is still greater than zero and I don't know how one can turn it into zero. Can someone please clarify?

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  • $\begingroup$ It means: for every $\epsilon > 0$. $\endgroup$ – Mauro ALLEGRANZA Oct 30 '17 at 7:24
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    $\begingroup$ Assume not, i.e. $\text {diam}(\overline E) > \text {diam}(E)$. This means that tehre is an $\epsilon_0 >0$ such that: $\text {diam}(\overline E) = \text {diam}(E) +\epsilon_0$. $\endgroup$ – Mauro ALLEGRANZA Oct 30 '17 at 7:25
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    $\begingroup$ But thus, e.g.: $\text {diam} (\overline E) > \text {diam}(E) + \dfrac {\epsilon_0} {2}$. $\endgroup$ – Mauro ALLEGRANZA Oct 30 '17 at 7:28
  • $\begingroup$ If a number that is known to be non negative is smaller than any positive number, then..? $\endgroup$ – Alvin Lepik Oct 30 '17 at 9:11
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We want to show that if $\forall \epsilon > 0, a \leq b + \epsilon$, then $a \leq b$.

Suppose $a > b$, then we let $\epsilon = \frac{a-b}{2}>0$

then we have $$a \leq b + \frac{a-b}{2}=\frac{a+b}{2}$$

Simplifying, we have $$2a \leq a+b$$

and hence $$a \leq b$$

but we have assumed that $a>b$ which is a contradiction since we get $a < a$.

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  • $\begingroup$ I like this answer. It does not provide an intuitive explanation but it has to convince anybody doubting this step by using algebra only. $\endgroup$ – Dirk Oct 30 '17 at 8:29
  • $\begingroup$ " contradiction of $a < a$" ? $\endgroup$ – Mauro ALLEGRANZA Oct 30 '17 at 9:53
  • $\begingroup$ my english isn't so good, let me try to rephrase it. $\endgroup$ – Siong Thye Goh Oct 30 '17 at 16:50
  • $\begingroup$ @SiongThyeGoh Thank you. Did you mean "a < b" in your last line? $\endgroup$ – user1691278 Oct 30 '17 at 23:17
  • $\begingroup$ $a \leq b$ and $b <a$ gives me $a < a$ which is a contradiction. $\endgroup$ – Siong Thye Goh Oct 31 '17 at 1:41
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Consider the following similar question: "Suppose the nonnegative real $r$ is less than or equal to $\epsilon$, for every positive $\epsilon$. Then $r = 0$."

Let $\epsilon = \frac{1}{2}$ to demonstrate that $r \leq \frac{1}{2}$. Similarly, $\epsilon = \frac{1}{4}$ demonstrates that $r \leq \frac{1}{4}$; and so on, to demonstrate that $r \leq \frac{1}{2^n}$ for each $n$. There's only one nonnegative number with that property.

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Specifically, the phrase "Since $\epsilon>0$ is arbitrary, [...]" is a short way to write "Since our argument up until now is true for any $\epsilon>0$, so is the final conclusion in the previous sentence, and therefore [...]".

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  1. If something is true for an arbitrary member of a set, it is true for all members of a set. E.g. Let $A, B$ be sets. Let $a\in A$ be arbitrary. Suppose this implies that $a\in B$. Then "no matter which $a$ you choose, $a$ is still in $B$" - in other words, "for all $a\in A$, $a\in B$".

Now we can move on to the important part :)

Suppose $a\leq b+\varepsilon$ for an arbitrary $\varepsilon > 0$. As I've explained above, this means that $a\leq b+\varepsilon$ for all $\varepsilon > 0$.

Suppose, with a view to contradiction, that $a \not\leq b$. Then $a > b$. This means $\frac{a-b}{2} > 0$, but now if we set $\varepsilon = \frac{a-b}{2}$, we see that $a > b +\varepsilon$, a contradiction.

This shows that $a < b$, as required.

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It means that it's true for any $\epsilon>0$. The way it goes away is because we know that $\Delta d = \operatorname{diam}(\overline E)-\operatorname{diam}(E) \le \epsilon$.

Now if $\Delta d>0$ we would have that $\Delta d/2>0$ so since $\Delta d>\epsilon$ for any $\epsilon>0$ we would have that it would be true for $\epsilon = \Delta d/2>0$ so we would have that $\Delta d \le \Delta d/2$ that is $\Delta d/2\le 0$ which means that $Delta d\le 0$ (which contradicts our assumptions).

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The term arbitrary means that you can choose any $\epsilon > 0$ without any restrictions. You could choose $\epsilon$ to be 4, 1, 0.01, 0.00001, etc. Since there are no restrictions on $\epsilon$, the only way to ensure that $a < b + \epsilon$ is to take the infimum of $\epsilon$, i.e., 0. That way if I choose any $a$ such that $a < b$, my $a$ will always be less than your $b + \epsilon$ no matter which $\epsilon$ you choose.

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