I thought, is this really that simple? Or am I missing a piece? This is my proof: $f(\{x\})=\{f(x)\}$.

Look at $f(\{x\})$. By definition, $f(\{x\})=\{f(a)|a \in \{x\}\}$, and therefore $f(\{x\})=\{f(x)\}$.

  • $\begingroup$ Notice that you use $x$ in two meanings. Once it is the element for which you proof the result, in the second occurrence I would prefer $f(\{x\})=\{f(y); y\in\{x\}$.\\Using the same symbol in two meanings might be confusing and it is good to avoid such thing. $\endgroup$ Sep 16, 2011 at 8:06
  • $\begingroup$ BTW proof-theory is not a good tag for this question meta.math.stackexchange.com/questions/1843/… $\endgroup$ Sep 16, 2011 at 8:13
  • 3
    $\begingroup$ @Martin: This is a setting in which I very much prefer the notation $f[\{x\}]$; the result in question is then $f[\{x\}]=\{f(x)\}$, which is trivially true from the definition of $f[\cdot ]$. $\endgroup$ Sep 16, 2011 at 8:27
  • 1
    $\begingroup$ @Martin: I thought that it was probably something like that just from the context. My comment wasn’t really aimed at you; yours just gave me a convenient peg on which to hang it, so to speak, since it was another example of not using the same symbol for two purposes at once. $\endgroup$ Sep 16, 2011 at 8:45
  • 1
    $\begingroup$ In addition to $f[\cdot]$, there's also the problematic-with-analysts notation of $f''\cdot$. $\endgroup$
    – Asaf Karagila
    Sep 16, 2011 at 13:24

2 Answers 2


The definition of a function is a set of ordered pairs, such that if for some $x$ we have $\langle x,y\rangle,\langle x,z\rangle\in f$ then $y=z$.

That is to say, $f$ is a binary relation and for every $x$ there is at most one element such that $\langle x,y\rangle\in f$. We call often say that $y$ is $f(x)$.

Now, what is $f(A)$ when $A\subseteq\operatorname{Dom}(f)$? It is the set $\{f(x)\mid x\in A\}$. Let $A=\{x\}$, we have that $$f(\{x\}) = \Big\lbrace f(a)\mid a\in\{x\}\Big\rbrace$$

Since $a\in\{x\}\iff a=x$ we have that $f(\{x\}) = \{f(x)\}$ as wanted.


This is basically the same proof as yours, I just rewrote it more formally/at lower level.

To prove that $f(\{x\})=\{f(x)\}$ we need to show that: $$z\in f(\{x\}) \Leftrightarrow z\in \{f(x)\}.$$

This can be shown as follows:

$z\in f(\{x\})$ $\Leftrightarrow$ $(\exists a\in\{x\}) z=f(a)$ $\Leftrightarrow$ $z=f(x)$ $\Leftrightarrow$ $z\in\{f(x)\}$

(You might want to think for a bit why the equivalences I wrote above are true - in case you want to make a really detailed proof. I think that most of the details are explained nicely in Asaf's answer.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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