I understand that a conditional statement is said to be vacuously true in the case that its antecedent is false, in which case the conditional statement is, itself, true. However, what about the following:

Let $A$ and $B$ be sets such that $A \subseteq B$. Let $f: A \to \mathbb{R}$ be a map. Let $b \in B$ such that $f(b)$ is not defined. Fix an arbitrary $\varepsilon > 0$ and consider the following statement: $$ f(b) < \varepsilon. $$

Is this true or false? The more I think about it, the more ambiguous it seems. After all, if it is true then its negation $ f(b) \geq \varepsilon $ should be false, yet this makes no more sense than for $ f(b) < \varepsilon $ to be false in the case that $f(b)$ is undefined. Is there another term for this besides vacuous truth?

Edit: The context in which this question arises is if a function can be considered continuous at a point at which it is not defined. Let $X$ and $Y$ be metric spaces. Let $A \subseteq X$. Let $a \in A$. Let $f : A \to Y$ be a map. Then, define the phrase "$f$ is continuous at $a$'' to mean the following: $$ \forall \varepsilon > 0, \hspace{1mm} \exists\hspace{1mm} \delta > 0 \ni \hspace{4mm} x\in N_\delta^X(a)\cap A \implies d_Y (f(x), f(a)) < \varepsilon $$ where $N_\delta^X(a)$ denotes a neighborhood in the metric space $X$ of radius $\delta$, centered at $a$, and $d_Y$ denotes the metric of $Y$. So, if we apply this to an $x\in X \setminus A$, $f(x)$ is undefined and $x \in N_\delta^X(a)\cap A$ is false. Does it then follow that the conditional statement in the definition of continuity at $x$ is vacuously true. Hence, $f$ is continuous at at point $x$ for which $f(x)$ is undefined.

So, in the original question, $f$ is like $d_Y$ and $b$ is like the pair $(f(x), f(a)) \in Y \times Y$. Analytic definitions aside, the heart of the question is this: If an implication is characterized by a false antecedent and a consequent which is neither true nor false. Is the implication itself true?

Thank you.

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    $\begingroup$ A statement of the form "For all $x$ such that a certain condition holds, ..." is vacuously true iff the certain condition holds for no $x$. $\endgroup$ – Evan Aad Jul 19 '19 at 14:59
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    $\begingroup$ if $f(b)$ not defined, it isn't a real number, so inequalities of any kind mentioning it don't make sense. $\endgroup$ – coffeemath Jul 19 '19 at 15:00
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    $\begingroup$ The "statement" $f(b)<\epsilon$ is not true and is not false. It is a "statement" without sense because $f(b)$ is not defined. Actually you could also say that it is no statement at all, so that it cannot be labeled with "true" or "false". $\endgroup$ – drhab Jul 19 '19 at 15:00
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    $\begingroup$ @coffeemath In the context of set theory, it is (when fully unfolded into set theoretic primitives) a valid formula regardless of whether it makes "intuitive" sense or not. $\endgroup$ – Derek Elkins left SE Jul 19 '19 at 23:06
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    $\begingroup$ @DerekElkins How is it a valid formula if one of its terms is not defined? In any interpretation of a valid formula, don't each of its terms need to be assigned to a member of the chosen universe? $\endgroup$ – coffeemath Jul 20 '19 at 3:15

In set-theoretic foundations, the statement is false. The problem is the distinction between logical functions and set-theoretic functions. If $f$ is a (unary) logical function (i.e. a function symbol in our theory), then $f(x)$ is always defined.1 To this end, your question wouldn't make sense. However, it is clear that you intend a set-theoretic function (at least assuming you are using a set-theoretic foundation) and so a function is a set, i.e. an individual of set theory. To this end, $f(x)$ is not a term at all, instead the formula $f(x)=y$ is shorthand for $(x,y)\in f$. Your formula: $f(b)<\varepsilon$ expands to $\exists y.(b,y)\in f\land y<\varepsilon$ which is simply false when $b\notin A$.

1 There are some logics that have an inherent notion of a not fully defined function, but those are non-standard.

  • $\begingroup$ Ok I am (luckily) somewhat familiar with the set theoretic notion of a function, although I haven't had much practice with it (if $f$ maps $A$ into $B$ then $f$ is a subset of $A \times B$, is it not?). I like the general direction of this answer quite a bit. Here is a similar but different question. If $f: A\to B$ is a map then $f^{-1}(f(C)) \supseteq C$ in general for $C \subseteq A$. I would like to prove that this holds with equality (that $f^{-1}(f(C)) \supseteq C$) in the case $C=A$. Let $x\in f^{-1}(f(C))$. Certainly, $f(x) \in f(C)$. (continued) $\endgroup$ – Thomas Winckelman Jul 20 '19 at 0:03
  • $\begingroup$ My first thought was to use contraposition and try to prove that $x \not\in A \implies f(x) \not\in f(A)$. But then I got stuck on the meaning of $f(x) \in f(A)$ when $x$ is not an element of the domain of $f$. With your answer though, would the meaning be as follows? : $x \not\in A \implies (x,b) \not \in A \times B$ for any possible element $b\in B$ of any set. In other words, if $f(A)$ is the set $f$ (which I’m not 100% sure of), then $ f(x) \not\in f(A)$. Is that the idea? $\endgroup$ – Thomas Winckelman Jul 20 '19 at 0:03
  • $\begingroup$ (oops, I mean to sat $f^{-1}(f(C) \subseteq C$ in the case $C = A$, sorry) $\endgroup$ – Thomas Winckelman Jul 20 '19 at 0:38
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    $\begingroup$ Using the contrapositive here is not particularly necessary here, and I would generally recommend avoiding indirect proofs when possible, but anyway, your formula "really means" $x\notin A\implies\exists y.(x,y)\in f\land y\notin f(A)$. Taking instead $f(x)\in f(A)\implies x\in A$ to prove this really just involves taking the $\exists y.(x,y)\in f$ part and using the fact that, by definition, $f\subseteq A\times B$ which means $(z,w)\in f\implies z\in A\land w\in B$, from which $x\in A$ immediately follows. $\endgroup$ – Derek Elkins left SE Jul 20 '19 at 11:16
  • $\begingroup$ An excellent answer. You managed whether by coincidence or by wisdom to address the root of my problem. Thank you $\endgroup$ – Thomas Winckelman Jul 20 '19 at 14:57

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