I am having difficulty in explaining the terminology "defined" to the students I am assisting. Here is the sentence: If a real-valued function $f$ is defined and continuous on the closed interval $[a,b]$ in the real line, then $f$ is bounded on $[a,b]$. Can I have some thoughts on how to explain the word "defined" used in the sentence? Thank you.

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    $\begingroup$ $f$ is defined on the interval $[a, b],$ means that we know $f$ (either its value, its expression, or how to compute it) for every $x \in [a, b].$ Outside this interval, we know nothing about $f.$ $\endgroup$
    – user2468
    Jul 5, 2012 at 17:26
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    $\begingroup$ ...or, put another way, the function has no discontinuities within the closed interval being considered. $\endgroup$ Jul 5, 2012 at 17:27
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    $\begingroup$ "$f$ is defined on the closed interval $[a,b]$" means the 'domain' of $f$ is $[a,b]$ (Note that 'domain','codomain' are necessary ingredients to describe any function). $\endgroup$
    – pritam
    Jul 5, 2012 at 17:28
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    $\begingroup$ Great. Thanks Matt, J.D., J.M. and Pritam. Now I got an idea on how to explain it. $\endgroup$
    – Sandra
    Jul 5, 2012 at 17:31

2 Answers 2


What age students? If it's just a precalculus or calculus course, I would just give examples of a nice looking formula that "isn't defined" on all of an interval, e.g. $\log(x)$ on [-.5, 2] or $1/x$ on [-1, 1].

If it's an analysis course, I would interpret the word defined in this sentence as saying, "there's some function $f$, taking values in $\mathbb{R}$, whose domain is a subset of $\mathbb{R}$, and whatever the domain is, definitely it includes the closed interval $[a, b]$."

In general the mathematician's notion of "domain" is not the same as the nebulous notion that's taught in the precalculus/calculus sequence, and this is one of the few cases where I agree with those who wish we had more mathematical precision in those course. (Often "domain" means something like "I wrote down a formula, but my formula doesn't make sense everywhere. Tell me where it does make sense," which I hate, especially because students are so apt to confuse functions with formulas representing functions.)


I support the point made by countinghaus that confusing a function with a formula representing a function is a really common error. Later on when things are complicated, you need to be able to think very clearly about these things. However, I also guess from other comments made that there is a bit of a fuzzy notion present in precalculus or basic calculus courses along the lines of 'the set of real numbers at which this expression can be evaluated to give another real number'....?

Anyhow, if we are to be proper and mathematical about this, it seems to me that the issue with understanding what it means for a function to be defined on a certain set is with whatever definition of `function' you are using. The way I was taught, functions are things that have domains. I agree with pritam; It's just something that's included. For example, a measure space is actually three things all interacting in a certain way: a set, a sigma algebra on that set and a measure on that sigma algebra. Given the sigma algebra, you could recover the "ground set" by taking the union of all the sets in the sigma-algebra.

A function is a domain $A$ and a codomain $B$ and a subset $f \subset A\times B$ with the property that if $(x,y)$ and $(x,y')$ are both in $f$, then $y=y'$ and that for every $x \in A$ there is some $y \in B$ such that $(x,y) \in f$. We write $f : A \to B$. If $(x,y) \in f$, we write $f(x) = y$. We may say, for any set $S \subset A$ that $f$ is defined on $S$.


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