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I've seen continuity in R defined in the two following ways. Namely

$$\forall c \in X, \forall \epsilon > 0 , \exists \delta>0, ((x\in X \land |c-x|<\delta) \implies |f(c)-f(x)| < \epsilon) $$

$$\forall c \in X, \forall \epsilon > 0 , \exists \delta>0, \forall x\in X (|c-x|<\delta \implies |f(c)-f(x)| < \epsilon) $$

Thus (letting $A = |c-x|<\delta$ and $ B= |f(c)-f(x)| < \epsilon$) it must be that for statements $A,B$, the following are equivalent.

$$\forall x \in X,(A\implies B) $$ $$(x \in X \land A) \implies B$$

I'd like to prove this formally, but I'm not sure how. I tried calling both $x\in X$ and $\forall x \in X$ statement $C$, and making a truth table. I found that $$C \land (A\implies B) \equiv (C \land A) \implies B$$ iff $C$ is true. I'm quite confused; do quantifiers like $\forall x \in X$ even have truth values?

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  • $\begingroup$ Quantifiers do not have truth values. I think in the second statement the brackets have been omitted and that is the only difference (but the statement is to be interpreted in the same way). It looks like your $\epsilon$ and $\delta$ are not quantified correctly? $\endgroup$ – AnyAD Apr 21 '17 at 0:40
  • $\begingroup$ @any the 2nd definition was from wikipedia: en.wikipedia.org/wiki/Uniform_continuity. The first was Elementary Analysis The Theory of Calculus, Ross, 2e page 124. $\endgroup$ – Evan Rosica Apr 21 '17 at 0:48
  • $\begingroup$ Notice: It is $\ldots\forall\varepsilon{>}0~\exists \delta{>}0\dots$ Note also the difference between continuity and uniform continuity. $\endgroup$ – Graham Kemp Apr 21 '17 at 0:54
  • $\begingroup$ @Evan Rosica So it should be for all epsilon there exists delta? Unless this is special type of continuity (eg uniform)? $\endgroup$ – AnyAD Apr 21 '17 at 0:55
  • $\begingroup$ @Any yes, it should be $\forall \epsilon>0, \exists \delta>0 $. good catch . That said it shouldn't affect the question. That was just context. In uniform continuity the $\forall c \in X$ would move after the $\exists \delta $ $\endgroup$ – Evan Rosica Apr 21 '17 at 0:59
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No, quantifiers do not have truth values, on their own.   However a restricted quantifier has equivalent forms, and we can use universal generalisation.

$$\begin{align}(\forall x\in X) (A(x)\,\to\, B(x)) ~&\iff (\forall x)\Big(x\in X\to\big(A(x)\to B(x)\big)\Big) \\[1ex] ~&\iff (\forall x)\big((x\in X\,\wedge\, A(x))\,\to\, B(x)\big)\\[1ex] &\;\implies (x_1\in X\,\wedge\, A(x_1))\;\to\; B(x_1)\end{align}$$

So the formal definition is: $~\forall c {\in} X~ \forall \varepsilon {>} 0 ~ \exists \delta{>}0 ~ \forall x{\in} X ~\big(\lvert c-x\rvert<\delta ~\to~ \lvert f(c)-f(x)\rvert < \varepsilon\big)~$ where all variables are bound.

Infering a generalised form: $\forall c {\in} X~ \forall \varepsilon {>} 0 ~ \exists \delta{>}0 ~ \big(\big(x\in X~\wedge~\lvert c-x\rvert<\delta\big) ~\to~ \lvert f(c)-f(x)\rvert < \varepsilon\big)$ where the arbitrary variable, $x$, is free.


Note:

  • The first form is the stronger uniform continuity, and states that there exists a $\delta$ for all $x$, which satisfies the predicate.

  • The second form is weaker continuity; it states that for each arbitrary value, $x$, there is some $\delta$ value that satisfies the predicate.

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  • $\begingroup$ The first form is not uniform continuity. The first form was Theorem 17.2 from Ross. Uniform continuity is Defn 19.1 from Ross, and states: $$\forall \epsilon >0, \exists \delta >0, \forall x,y\in S |x-y| < \delta \implies |(f(x)-f(y)|< \epsilon $$ $\endgroup$ – Evan Rosica Apr 21 '17 at 1:08
  • $\begingroup$ The difference between the first definition I gave in my original post and the definition of uniform continuity above is that the $\forall c$ moves after the $\delta$ (indicating $\delta$ no longer depends on $c$), and Ross renames $c=y$, since the roles of $x,y$ are symmetric. $\endgroup$ – Evan Rosica Apr 21 '17 at 1:24

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