I was thinking about the limit definition and I'm having an hard time understanding why it isn't formulated in terms of logical conjunction instead of implication, that is: let $f:A\subseteq \mathbb{R}\to\mathbb{R}$ and $x_0$ a limit point of $A$, why the definition of limit isn't formulated like $$\forall \epsilon>0\exists \delta_\epsilon >0 \ \text{s.t.} \forall x\in A, \ 0<|x-x_0|<\delta_\epsilon \land |f(x)-l|<\epsilon$$ Instead of the well known $$\forall \epsilon>0\exists \delta_\epsilon >0 \ \text{s.t.} \ \text{s.t.} \forall x\in A, 0<|x-x_0|<\delta_\epsilon \implies|f(x)-l|<\epsilon$$ I know that intuitively the implication describes a situation of cause effect, so when I imagine on how I would define the situation of limit I surely want that "the fact that I'm going near a limit point causes the fact that the function is near to its limit value" and so I get why there is the implication, but I can't understand why the logical conjunction can't describe the same situation since when I use $\land$ in the proposition "$0<|x-x_0|<\delta_\epsilon \land |f(x)-l|<\epsilon$" I want that, because of the definition of logical conjunction, both the condition $0<|x-x_0|<\delta_\epsilon$ and $|f(x)-l|<\epsilon$ are satisfied and so I still see a situation similar to the cause effect given by the implication and it still seems plausible to me defining limit this way. Can someone help me understand why this doesn't work? Thank you.

  • $\begingroup$ Both formulations are wrong, since you're missing the quantifier on $x$. $\endgroup$ Commented Apr 10, 2021 at 2:31
  • $\begingroup$ @EricWofsey: Thank you, I edited my question hoping that now it is well asked. $\endgroup$
    – Gwyn
    Commented Apr 10, 2021 at 2:35
  • $\begingroup$ Logically difference between default limit definition and your suggestion is same as with $\neg A \lor B$ and $A \land B$. $\endgroup$
    – zkutch
    Commented Apr 10, 2021 at 2:59
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    $\begingroup$ Is it really the case, given $\epsilon$ and $\delta_\epsilon$, that every $x$ is within $\delta_\epsilon$ of $x_0$? If not, then we don't want "$0<\vert x-x_0\vert<\delta_\epsilon$" as an outright claim, as opposed to hypothesis, in the part of the definition inside the quantifiers. $\endgroup$ Commented Apr 10, 2021 at 3:24
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    $\begingroup$ Logical implication has nothing to do with cause and effect. If the moon is made of cheese, then the sun is as cold as ice. $\endgroup$
    – user21820
    Commented Apr 11, 2021 at 12:24

1 Answer 1


Let's simplify a bit and compare the statements $$(*)\quad\forall x(P(x)\wedge Q(x))$$ and $$(\dagger)\quad\forall x(P(x)\rightarrow Q(x)).$$ (Basically, I'm "suppressing" the complicated $\epsilon/\delta$ stuff to focus on the core logical structure.) Statement $(\dagger)$ says that every $x$ satisfying a certain condition, $P$, has a certain further property, $Q$. By contrast, $(*)$ asserts that every $x$ whatsoever has property $P$ and has property $Q$. Intuitively, $(*)$-statements are extremely rarely true, while $(\dagger)$-statements are quite common (e.g. "Every multiple of $4$ is even" is a $(\dagger)$-statement: take $P$ to be "is a multiple of $4$" and $Q$ to be "is even").

Think of a $(\dagger)$-statement as saying "I only care about the $x$s with property $P$, and all of those $x$s have property $Q$." That's exactly the sort of thing that's going on in the limit definition: we only care about the $x$s within a certain distance of $x_0$ (and which are not $x_0$ itself), and all of those $x$s have a certain additional property. This "attention-restriction" may feel like a conjunction, but it very much isn't; thinking about the "Every multiple of $4$ is even" example will, I think, help clarify this.

EDIT: If you're already comfortable thinking in terms of bounded quantifiers (e.g. "$\forall x\in U(Q(x))$"), the discussion here may be useful.

  • $\begingroup$ Thanks for giving this in good detail highlighting the key aspects in bold. +1 $\endgroup$
    – Paramanand Singh
    Commented Apr 13, 2021 at 3:02

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