Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose I have a function $f: X \rightarrow X$, a finite set $S\subset X$ and $|S'| \neq |S|$ for $f(S) = S'$. Is there a special name for such a function $f$?

share|cite|improve this question
Not $1-1. \ \ \ \ $ – P.. Mar 2 '13 at 13:59
I thought this was going to be about infinite sets! Does anyone know the name for it in that case? – Tara B Mar 2 '13 at 14:02
Ah, no, $X$ is finite as well – Confusion Mar 2 '13 at 14:03
Yes, that doesn't really matter. What I thought you were going to be asking for (from the title) is the name for a function $f: X\to X$ where $|f(X)|<|X|$. If $X$ is infinite, that is a much stronger property than 'not injective'. – Tara B Mar 2 '13 at 14:06
I realised that the comments on another question are a bad place to ask a question, so I've now asked it here. – Tara B Mar 2 '13 at 14:42
up vote 2 down vote accepted

If you want to specifically mention that the image of $S$ under $f$ is smaller than $S$, you could say that $f$ is "not injective on $S$". Of course $f$ is also not injective at all, as the other answers say, but that is weaker.

share|cite|improve this answer
+1, good answer, in fact the formulation of the question appears to require a name that refers to both $f$ and $S$. – Andreas Caranti Mar 2 '13 at 14:10
@AndreasCaranti: Thanks. I do think this is probably what the OP was looking for. – Tara B Mar 2 '13 at 14:11

Then you must have $|S'| < |S|$, and you are talking of a function which is not injective.

share|cite|improve this answer

Yes: in that case it must be that $|S'|<|S|$, and therefore $f$ is not injective.

share|cite|improve this answer
Not literally the same answers, but we came close ;-) – Andreas Caranti Mar 2 '13 at 14:00
@Andreas: Hard to get much closer! – Brian M. Scott Mar 2 '13 at 14:00
So given my edit, 'not injective and not surjective', but that's all – Confusion Mar 2 '13 at 14:01
@Confusion: If $X$ is infinite, your function could be surjective and still have the stated property. E.g., $$f:\Bbb N\to\Bbb N:n\mapsto\left\lfloor\frac{n}2\right\rfloor\;.$$ – Brian M. Scott Mar 2 '13 at 14:02

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.