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I came across this problem while trying to work out the symbolic expression of a particular definition, and have been stumped since.

Take this definition of continuity

A function $f\colon D \to R \,$ is continuous at $c \in D \,$ if for every $\epsilon >0$ there is a $\delta>0$ such that $|f(x)-f(c)|<\epsilon \,$ if $\, |x-c|<\delta$ and $x \in D$ .

However, in trying to express this in symbolic terms, I run into a problem. The implication part of this definition, i.e. "$|f(x)-f(c)|<\epsilon \,$ if $\, |x-c|<\delta$ and $x \in D \,$", looks to me like $ x \in D \ \land \ |x-c|<\delta \implies |f(x)-f(c)|<\epsilon$ in symbolic terms. This is considered wrong however, apparently the "and" in my definition isn't a conjunction, but just a non-connective kind of "and". What exactly is this "and" then, and how do I distinguish between logical connective and's and non-connective and's?

It's also worth noting that $x \in D$ in the implication should be translated to symbolic terms $\forall x \in D$, according to my book. Why? The resulting statement would be

A function $f\colon D \to R \,$ is continuous at $c \in D \,$ if for every $\epsilon >0$ there is a $\delta>0$ such that if $\, |x-c|<\delta$ and for all $x \in D$, then $|f(x)-f(c)|<\epsilon \,$

...which sounds plain ridiculous and incomprehensible.

Could anybody offer some insight?

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    $\begingroup$ What do you mean by "is considered wrong"? Who considers it wrong? It looks correct to me (except that you should have parentheses: $(x \in D \ \land \ |x-c|<\delta) \implies |f(x)-f(c)|<\epsilon$). $\endgroup$ Commented Sep 10, 2017 at 22:39
  • $\begingroup$ The lack of parentheses is disturbing. Even more so that is socially acceptable to write that way. $\endgroup$
    – Git Gud
    Commented Sep 10, 2017 at 23:05
  • $\begingroup$ @GitGud That's the way my textbook taught me, I thought it was an acceptable way of writing these expressions. $\endgroup$
    – Ius Klesar
    Commented Sep 10, 2017 at 23:10
  • $\begingroup$ Socially it is. Technically it's very wrong. Most calculus/analysis text books will make such offenses, don't worry. $\endgroup$
    – Git Gud
    Commented Sep 10, 2017 at 23:48
  • $\begingroup$ The accepted convention is that the conditional has lower precedence that conjunction, and so $A\wedge B\to C$ is read as $(A\wedge B)\to C$. However, it is best practice to use parenthesis, because people do get confused about operation order when you don't make it very clear. $\endgroup$ Commented Sep 11, 2017 at 21:54

2 Answers 2

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Anyone who claims that it is wrong to interpret this "and" as a conjunction is simply wrong. There is absolutely nothing wrong with translating $$\text{"$|f(x)-f(c)|<\epsilon \,$ if $\, |x-c|<\delta$ and $x \in D \,$"}$$ as $$( x \in D \ \land \ |x-c|<\delta) \implies |f(x)-f(c)|<\epsilon$$ (though note the parentheses which you left out). The meaning of this "and" is exactly the usual conjunctive meaning and it is not a different kind of "and". Note, though, that in the context of the definition of continuity there is an implicit universal quantifier on the variable $x$ (the statement must be true for all $x$), so enlarging the context a little bit you could also translate it as $$\forall x(( x \in D \ \land \ |x-c|<\delta) \implies |f(x)-f(c)|<\epsilon).$$

When your book suggests to interpret it as $\forall x\in D$, they mean to put it at the start of the conditional, not in the middle as you did. In doing so, they are rephrasing the statement as a logically equivalent statement which leaves out the "and" (and thus the conjunction). Specifically, they mean $$\text{"for all $x\in D$, if $\, |x-c|<\delta$, then $|f(x)-f(c)|<\epsilon$"}$$ or in symbols, $$\forall x\in D\ (|x-c|<\delta \implies |f(x)-f(c)|<\epsilon).$$

Note that this statement is logically equivalent to your version with $\forall x$ at the start, so both are equally correct.

(As a final sidenote, there are uses of "and" which are not conjunctions. For instance, if I say "the two largest prime numbers less than $50$ are $47$ and $43$" my "and" isn't really a conjunction, and I am using a construction of the English language that is fairly complicated to transform into formal logic. This is totally unrelated to what is going on in your situation though.)

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  • $\begingroup$ What a great and thorough answer, thank you. That clarified a lot, and reassured me that this is the right way to go about it. Just wondering though, what would the negation be of the "for all" symbolic definition that you gave as a follow-up to mine - the one with the conjunction in parentheses? $\endgroup$
    – Ius Klesar
    Commented Sep 10, 2017 at 23:08
  • $\begingroup$ I'm not sure what you're looking for exactly: you can just stick a negation symbol in front, like $\neg\forall x(( x \in D \ \land \ |x-c|<\delta) \implies |f(x)-f(c)|<\epsilon)$. $\endgroup$ Commented Sep 10, 2017 at 23:27
  • $\begingroup$ Just as an aside -- it's usually best to imagine all quantifiers as having a specified domain, whether or not you write it explicitly. Here, it looks like you're talking about real numbers, so you could imagine Eric's statement as $\forall x\in \mathbb R (x\in D \land |x-c|<\delta) \Rightarrow |f(x)-f(x)<\epsilon)$. Since (assumedly) $D\subset\mathbb R$ you can shorten this. But an uninterpreted $\forall x$ makes me twitch -- the rest of the statement is neither true nor false when $x$ is "of the wrong kind" -- say, a set of sets of sets of sets of ... of reals? $\endgroup$ Commented Sep 11, 2017 at 1:56
  • $\begingroup$ @EricWofsey Eric, I just cross-checked this with my textbook, but it says using a conjunction like I suggested is wrong because, in a statement like $\forall x ((x \in D \land |x-c|<\delta) \implies (|f(x)-f(c)|< \epsilon$ the conjunction operator would mean that I'm actually saying "all real numbers are in D and satisfy this implication", when clearly not all real numbers have to be in the function's domain set $D$ (in the case of positive square-root functions, which cover only the positive real numbers and 0). In any case, it's not the same statement we started with (check the original). $\endgroup$
    – Ius Klesar
    Commented Sep 11, 2017 at 14:31
  • $\begingroup$ That would be the meaning if you parenthesized it the other way, like $\forall x (x \in D \land (|x-c|<\delta \implies |f(x)-f(c)|< \epsilon))$. The error isn't translating it as a conjunction but rather just parenthesizing it wrong. $\endgroup$ Commented Sep 11, 2017 at 14:58
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However, in trying to express this in symbolic terms, I run into a problem. The implication part of this definition, i.e. "$|f(x)−f(c)|<ϵ$ if $|x−c|<δ$ and $x∈D$", looks to me like $x∈D ∧ |x−c|<δ⟹|f(x)−f(c)|<ϵ$ in symbolic terms. This is considered wrong however,

It is not exactly wrong, but perhaps you should add parenthesis to be clear you have the correct form of $(A\wedge B)\to C$, rather than the completely different $A\wedge (B\to C)$.

apparently the "and" in my definition isn't a conjunction, but just a non-connective kind of "and". What exactly is this "and" then, and how do I distinguish between logical connective and's and non-connective and's? It's also worth noting that $x∈D$ in the implication should be translated to symbolic terms $∀x∈D$, according to my book. Why?

There is an implicit universal quantifier, because $x$ is not a particular constant value, but rather an arbitrary variable.

$$\forall x~\big((x\in D \wedge \lvert x−c\rvert <\delta)~\to~(\lvert f(x)−f(c)\rvert <\epsilon)\big)$$

Since $(A\wedge B)\to C$ is equivalent to $A\to (B\to C)$, this can also be written as : $$\forall x~\Big(x\in D \to \big((\lvert x−c\rvert <\delta)~\to~(\lvert f(x)−f(c)\rvert <\epsilon)\big)\Big)$$

And by recalling that: $\forall x~(x\in D\to P(x))$ is often abbreviated as $\forall x{\in}D~P(x)$

$$\forall x{\in}D~\big((\lvert x−c\rvert <\delta)~\to~(\lvert f(x)−f(c)\rvert <\epsilon)\big)$$

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  • $\begingroup$ "Since (A and B) -> C is equivalent to A -> (B -> C), this can also be written as..." How is the statement at the end of your post an example of A -> (B -> C)? I only see one implication symbol, and the only thing that changed is that you moved the "x in D" outside of the parentheses? $\endgroup$
    – Ius Klesar
    Commented Sep 10, 2017 at 23:27
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    $\begingroup$ @IusKlesar: $\forall x\in D (P)$ is an abbreviation for $\forall x (x\in D\implies P)$. $\endgroup$ Commented Sep 10, 2017 at 23:29
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    $\begingroup$ Indeed. Added the apparently-not-obvious extra step. $\endgroup$ Commented Sep 10, 2017 at 23:39
  • $\begingroup$ @GrahamKemp I've cross-checked this with my textbook, but it says using a conjunction like I suggested is wrong because, in a statement like $\forall x ((x \in D \land |x-c|<\delta) \implies (|f(x)-f(c)|< \epsilon)$ the conjunction operator would mean that I'm actually saying "all real numbers are in D $and$ satisfy this implication", when clearly not all real numbers have to be in the function's domain set $D$ (in the case of positive square-root functions, which cover only the positive real numbers and 0). $\endgroup$
    – Ius Klesar
    Commented Sep 11, 2017 at 14:29
  • $\begingroup$ @IusKlesar - NO; it reads: "if (a real numbers $x$ is in $D$ and satisfy this implication), then ..." and this holds for every real number. If $x \notin D$ then the concjunction is false and the conditional is true (a conditional $F \to ?$ is $T$, whatever is the value of $?$). $\endgroup$ Commented Sep 11, 2017 at 15:51

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