The question arises from the rewording of some theorems, by example continuity of a function $f$, we have that

$$f\text{ is continuous at }c\iff (\forall\epsilon>0,\exists\delta>0:|x-c|<\delta\implies|f(x)-f(c)|<\epsilon)\tag{1}$$

then I think I can reword this as

$$f\text{ is continuous at }c\iff ((|x-c|<\delta\implies|f(x)-f(c)|<\epsilon)\implies (\forall\epsilon>0,\exists\delta>0))$$

The last statement seem strange to read but the last implication seems logically equivalent to $(1)$. Anyway this is just an experiment.

My question is, there is some expression logically equivalent to "such that" that can be written with basic logical operators as $\lor$, $\land$, $\lnot$, $\iff$ and $\implies$? Thank you in advance.


No, you can't do that; that's not how quantifiers (or conditionals) work.

"$\forall \epsilon>0$" just means, "For all $\epsilon>0$." It doesn't assert anything - e.g. it doesn't make sense to say ""$\forall \epsilon>0$" is false". Similarly with "$\exists \delta>0$."

So an expression of the form "...$\implies (\forall \epsilon>0,\exists \delta>0)$" is meaningless: "$\implies$" connects two sentences.

The issues go further than this: quantifiers bind variables. The expression "$\forall x(x>5)$" is a sentence; the expression "$x>5$" is not, and the expression "$(x>5)\implies (\forall x)$" is really not.

  • $\begingroup$ Then the meaning of "such that" is irreducible to other logic operators, right? $\endgroup$ – Masacroso Oct 18 '16 at 3:33
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    $\begingroup$ @Masacroso I don't understand what you're asking. "Such that" is implicit in how quantifiers work - e.g. "$\exists x(P(x))$" means "There exists an $x$ such that $P(x)$." So I think the answer to your comment is "no?" But I don't really understand what you mean by "irreducible to other logic operators" . . . $\endgroup$ – Noah Schweber Oct 18 '16 at 3:34
  • $\begingroup$ Oh, Ok, I understand... "such that" is implicit to quantifiers, silly me. $\endgroup$ – Masacroso Oct 18 '16 at 3:35
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    $\begingroup$ (An anecdote: I was not too long ago giving a seminar talk with a certain older logician in the audience. He frequently fidgeted when I wrote, and I had no idea why. After the talk, someone came up to me and asked a question, and in the process of answering I wrote "$\exists x$ s.t." on the board; the older logician he got up and erased the "s.t." and gave me a short historical lecture on how that was redundant. :P) $\endgroup$ – Noah Schweber Oct 18 '16 at 3:36

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