Using "implies" to refer to material conditional

Question

Is it acceptable to translate the binary connective "$\let\ f\rightarrow$" into English with "implies"? I'm unsure because "implies" for me immediately brings to mind logical implication, but I've seen some places use it for the material conditional (including wikipedia, in the opening sentence of this article).

For example, does mathematical convention, in principle, permit the following formulation of the standard definition for functional continuity?

$f$ is continuous at $c$ if for any $\epsilon > 0 :$ there exists $\delta > 0:$ for any $x$ in Domain[$f$]$:$ $|x-c|<\delta$ implies $|f(x)-f(c)|<\epsilon$.

Answer 1

• $\huge\rightarrow$ (the material conditional) is a logical connective that operates on its antecedent $P$ and its consequent $Q$ to form the truth function $$P\rightarrow Q,$$ which is false precisely when $P$ is true but $Q$ false.

• $\huge\Rightarrow$ (implication) similarly operates on its antecedent $P$ and its consequent $Q;$ here, the result $$P\Rightarrow Q$$ lives in a specific (usually implicitly understood) interpretation/context and signifies that $P\rightarrow Q$ is true in it. Some ways to read $P\Rightarrow Q:$

  • If $P$ is true, then $Q$ is true.
  • $P$ being true is a sufficient condition for $Q$ to be true.
  • $P$ being true implies that $Q$ is true.
  • $P$ is true only if $Q$ is true.
  • $Q$ being true is a necessary condition for $P$ to be true.

  When $P\not\Rightarrow Q,$ then it must be that $P$ is true yet $Q$ false.

In the given formulation $$|x-c|<\delta\;\; \textbf{implies} \;\;|f(x)-f(c)|<\varepsilon,$$ “implies” is not the material conditional $\large\rightarrow\normalsize$ per se, but rather mathematical implication $\large\Rightarrow\normalsize;$ it analytically (from mathematical axioms and a given context) asserts that its right side can be derived from its left.

P.S. I distinguish between implication $\,\large\Rightarrow\,$ and logical implication $\,\large\vDash\,,$ which is often used to mean first-order implication, i.e., that $P\rightarrow Q$ is true regardless of interpretation.

P.P.S. Symbolic logic is an area rife with conflicting notation, terminology and even notions; my understanding is eclectically evolving.

Answer 2

Yes.

By the way, logical implication is material conditional. In logic only the forms of the arguments matter in order to deduce from something.

When you see $$ |x-c|<\delta\text{ implies }|f(x)-f(x)|>\epsilon $$ written in a proof, it is certainly an English version of the formal statement $$ |x-c|<\delta\to|f(x)-f(x)|>\epsilon. $$

Sometimes authors say at the beginning of their book that the proofs will be given in an informal manner. Informal means that English language will be used for better readability. In principle, those informal proofs could be made formal in, say, first-order logic.

Note. The logical connective $\to$ really contains what we mean by "implies". Indeed, $p\to q$ does what it is supposed to do: it permits us to infer $q$ from $p$ but nothing from $\neg p$.