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.
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.