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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 $f(\{x\})=\{f(x)\}$.

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    In addition to $f[\cdot]$, there's also the problematic-with-analysts notation of $f''\cdot$.2011-09-16

2 Answers 2

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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.)

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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.