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There is another post that addresses this but quickly fix the problem to be something in arthmetics, and in turn what it means for that arithematics problem to be well defined.

I have never encontered the definition of "well defined" in a textbook, only in questions where I am asked to show something is well defined

For example

Given a functional

$$L(f) = \int\limits_0^1 [f^2 - f] dx$$

where $f\in [0,1] \to \mathbb{R}$, $f(0) = 0$ and Lipschitz, show $L(f)$ is well defined

Example 2:

Let $d$ be a metric on $X$, and define the point to set distance

$$d(x,C) = \inf\{d(x,y): y \in C\}$$

where $C \subset X$ is a nonempty set

Show $d(x,C)$ is well defined

In general, given object $\phi = f$, whatever $\phi, f$ is, what does it mean to show that $\phi$ is well defined?

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  • It essentially means that it is a function, meaning that for each input, there is one and only one output. – Aweygan Jul 17 '16 at 23:24
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    This depends on context. But generally, to say that a string of symbols is well defined, means that the entity which the string of symbols is supposed to represent actually exists. You can't just write any symbol you want and expect it to mean something.

    For instance, in the context of R-integrability, $\displaystyle \int_0^1f$, where $f\colon ]0,1[\to \mathbb{R},, x \mapsto \dfrac{1}{x}$, isn't well defined because $f$ isn't bounded on its domain, in other words, the integral doesn't exist.

     – Git Gud Jul 17 '16 at 23:25
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    In your first example, getting down to the nitty-gritty, showing that $L$ is well defined requires you to prove that $f^2$, $f$ and $f^2-f$ are functions, $f^2-f$ is integrable on $[0,1]$ and that for all functions $g,h$ in the given conditions, $g=h\implies \int\limits_0^1 g^2 - g\mathrm dx=\int\limits_0^1 h^2 - h \mathrm dx$. In your second example, among other things, you need to prove that the infimum of that set actually exists, because, as you know, it's not always true on any given partial order that the infimum exists for any given set. – Git Gud Jul 17 '16 at 23:27
  • Hello Git God, is there any reference that actually defines "well defined", or does "well defined" operate more like "trivial" is used colloqiually – Shamisen Expert Jul 17 '16 at 23:32
  • See the entry "well defined" in MathWorld or in Wikipedia. – Pedro Jul 17 '16 at 23:33
  • @BarackUnchained There isn't a mathematical definition of well-definiteness, it's more of a meta-mathematical concept. The best one can do is as is shown in Pedro's links above. Edit: I am yet to become a god. – Git Gud Jul 17 '16 at 23:37

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There is no universal definition for "well defined." Here are three common uses of the term.

  1. $X, Y$ are sets, $\sim$ is an equivalence relation on $X$, $\tilde{X}$ is the set of subsets $S$ of $X$ having the property that $x, y \in S$ if and only if $x \sim y$. A typical element $S$ of $\tilde{X}$ is often written as $\overline{x}$, for some $x \in S$. Let $f: X \rightarrow Y$ be a function, and let $\overline{f}: \tilde X \rightarrow Y$ be the function given by $\overline{f}(\tilde x) = f(x)$.

When someone says that $\overline{f}$ is well defined, they mean that $f(x) = f(y)$ whenever $x \sim y$. Otherwise, $\overline{f}$ makes no sense as a function.

Example: $f: R \rightarrow S$ is a homomorphism of rings, $I$ is an ideal of $R$, and $\overline{f}: R/I \rightarrow S$ is given by the formula $\overline{f}(x+I) = f(x)$. Then $\overline{f}$ is well defined if and only if $I$ is contained in the kernel of $f$.

2 . $X, Y$ are sets, and $f:X \rightarrow Y$ is a function which is given by some formula. It is clear that the output of the formula is either undefined or else an element of $Y$.

When someone says that $f$ is well defined, they mean that the formula gives a valid output.

Example: $f: (-1,1) \rightarrow \mathbb{R}$, $$f(x) = 1 + x + x^2 + \cdots $$ Here $f$ is defined in terms of a limit, and to say that $f$ is well defined is to say that the limit exists.

Example: $f: (0,\infty) \rightarrow \mathbb{R}, x \mapsto \ln x$. This function is not well defined, because no one has given $\ln 0$ a definition as a real number.

3 . $X, Y$ are sets, and $f:X \rightarrow Y$ is a function which is given by some formula. The formula makes sense if we regard $f$ as a function from $X$ into a larger set $Z$ which contains $Y$.

When someone says that $f$ is well defined, what they mean that if $x \in X,$ then $f(x)$ by its formula is actually an element of the set $Y$.

Example: let $f: (-1,1) \cap \mathbb{Q} \rightarrow \mathbb{Q}$ be given by $$f(x) = 1 + x + x^2 + \cdots$$

Suppose we have already established that $f(x)$ is an element of $\mathbb{R}$. To say that $f$ is well defined, we would have to argue that $f(x)$ is actually in $\mathbb{Q}$.

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