Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a trivial vocabulary question.

It seems to me that "constant on every connected component of the domain" would be a reasonable definition of the term "componentwise constant", provided that there's not already some other concept conventionally denoted by that same expression.

The term, in quotes, gets a lot of google hits, suggesting it does conventionally refer to something. Is that it, or is it something else?

If something else, then what term should be used instead? If $f'=0$, then $f$ is $\cdots\cdots\cdots\cdots\cdots\cdots\cdots$. If $f'=g'$, then $f$ and $g$ differ by a $\cdots\cdots\cdots\cdots\cdots\cdots\cdots$ function.

Later note: Someone in the "comments" section below proposed "locally constant". I think that's mistaken, for reasons I explained there. The comment got five up-votes. Are those people confused or am I?

share|improve this question
I think what you are calling "componentwise constant" is usually called "locally constant." –  froggie May 6 '12 at 17:41
In my experience componentwise generally refers to the components of points in a Cartesian product; e.g., ordinary vector addition is defined componentwise. –  Brian M. Scott May 6 '12 at 17:42
@froggie : I think you're mistaken. The identity function on the rational numbers is constant on connected components of its domain (each connected component contains just one point) but is it "locally constant"? –  Michael Hardy May 7 '12 at 18:13
@Michael Hardy: you're absolutely right, locally constant is different than what you are saying. Silly mistake, but an easy one to make, because in many situations that come up in practice, the notions do coincide. Thanks for pointing this out! –  froggie May 7 '12 at 18:28
Whie locally constant is technically incorrect in general, I have to wonder what the situation is here that would make a separate term necessary... –  Harry Altman May 7 '12 at 20:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.