Assume f : D -> S function, where D, S are subsets of R, different from R.

We know that statement for any A from D, f(A) belongs to S is correct.

I wonder whether this statement is correct as well: for any A from R\D, f(A) does not belong to S

If it is correct, whether we can say, that for any A from R\D, f(A) belongs to R\S ?

  • 4
    $\begingroup$ The question should be more like this : If $A \in \mathbb R \setminus D$, what even does $f(A)$ mean? For example, what does $\ln(-1)$ mean? $\endgroup$ – Teresa Lisbon Apr 29 '18 at 8:47
  • $\begingroup$ I just want to know whether that statement is correct or not $\endgroup$ – Hayk Manukyan Apr 29 '18 at 8:58
  • $\begingroup$ for sure the value is undefined, but I hesitate in boolean value of the predicate which takes my mentioned statement as an argument $\endgroup$ – Hayk Manukyan Apr 29 '18 at 9:13

As астон вілла олоф мэллбэрг stated, the best way to think of this conceptually is that $f(a)$ is just nonsense when $a$ is not in the domain of $f$. As such your second statement has no truth-value one way or the other.

Set theory, usually considered the "foundation" of math, does, however, give such statements meaning (unfortunately, in my opinion). In particular, in set theory a function is usually modeled by a set of ordered pairs. Conceptually, the function $f:X\to Y$ is the set of ordered pairs $\{(x,y)\in X\times Y\mid f(x)=y\}$. $f(x)=y$ is then a shorthand for $(x,y)\in f$. Your first statement is then: $$\forall a\in D.\exists b\in S.(a,d)\in f$$ Your second statement is similarly: $$\forall a\in\mathbb R\setminus D.\exists b\in\mathbb R\setminus S.(a,b)\in f$$ $f$ not being defined outside of $D$ means there is no ordered pair $(a,b)\in f$ for which $a\notin D$. Thus the second statement is false: for every $a\in\mathbb R\setminus D$, $(a,b)\notin f$ no matter what $b$ is.

To see why I view the above as unfortunate, consider the real number $\pi$. Is $\pi(3)=7$? Your first reaction is probably, "$\pi$ is not a function. This makes no sense." But $\pi(3)=7$ means $(3,7)\in\pi$ and this is a completely meaningful set-theoretic statement. Your second reaction (or an alternative first reaction) is probably, "this is false", but whether it's true or false depends on the precise definition of $\pi$ (and $3$ and $7$ and ordered pairs).

For contrast, in most type theories the analogue to your second statement would be a type error and thus not a well-formed formula. It would thus not make sense to ask about its truth value anymore than it would make sense to ask about the truth value of $\neg\land )x$. Similarly for $\pi(3)=7$.


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