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For these two statements,
(1). There exists a positive real number $\dfrac {\varepsilon } {2}$ smaller than every positive real number $ {\varepsilon } $ .

(2). There is no positive real number smaller than every positive real number $ {\varepsilon } $ .

Questions :

  1. (1) contradicts with (2), which is wrong and which is right ?
  2. What I got confused to get this contradiction?
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    $\begingroup$ (1) doesn't make much sense. If you say "there exists...", then you implicitly assume that the number $\varepsilon/2$ is fixed, and hence so is $\varepsilon$, so you can't really say that this then holds for all $\varepsilon$. $\endgroup$ – Wojowu Nov 18 '16 at 8:51
  • $\begingroup$ In order for (1) to contradict (2), it would need to establish a positive number (not an expression) that is smaller than every positive real number. What number would that be? $\varepsilon/2$? That's not smaller than $\varepsilon/4$, though. $\endgroup$ – T.J. Gaffney Nov 18 '16 at 9:06
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The order of your logical symbols is important.

  1. For all $\epsilon > 0$ there exists a smaller real number. $$ \forall\epsilon>0\exists\delta>0 \text{ such that } \delta<\epsilon $$
  2. There does not exists a number that is smaller for all $\epsilon>0$. $$ \nexists \delta > 0 \forall \epsilon>0 \text{ such that } \delta<\epsilon $$

Note that these statements don't contradict eachother. Actually, they are the same statement. Call the second statement $A$. Then $A = (A^c)^c$.

\begin{align} A^c &= \exists \delta > 0 \forall \epsilon>0 \text{ such that } \delta<\epsilon. \\ (A^c)^c &= \forall\delta > 0 \exists\epsilon>0 \text{ such that } \delta>\epsilon \end{align} This is the same as statement 1 with $\epsilon$ and $\delta$ reversed.

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  • $\begingroup$ Then how to tell the order of logical symbols in a statement ? $\endgroup$ – iMath Nov 18 '16 at 9:36
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    $\begingroup$ @iMath The author should make that clear, by making the order of the sentence follow the order of the symbols: e.g. they should have written "For every $\epsilon>0$ there is a positive real smaller than $\epsilon$". In certain cases, a failure to do so is okay when there are other textual clues. For example, in the statement (1) above, "there exists a positive real number ${\epsilon\over 2}$" references $\epsilon$, so must come after the quantifier introducing $\epsilon$. This is, however, pretty bad writing, and I find the language of statement (1) fairly unclear. $\endgroup$ – Noah Schweber Nov 20 '16 at 0:05
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The subtlety here fooled me once to.

The key to understanding the contradiction is noticing what is fixed first.

For "There exists a positive real number $\dfrac {\varepsilon } {2}$ smaller than every positive real number $ {\varepsilon } $ " we first set $\epsilon $, any epsilon, and then say $\dfrac {\varepsilon } {2}$ is smaller, which is true.

For "There is no positive real number smaller than every positive real number $ {\varepsilon } $" we are saying that, there is no number that we can fix that will be smaller than any other. They are saying that whatever, for how many small, $x $ is, there will be a smaller $\epsilon $.

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