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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asked Sep 16, 2011 at 8:02
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$\begingroup$ 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. $\endgroup$– Martin SleziakSep 16, 2011 at 8:06
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$\begingroup$ BTW proof-theory is not a good tag for this question meta.math.stackexchange.com/questions/1843/… $\endgroup$– Martin SleziakSep 16, 2011 at 8:13
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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 $f[\cdot ]$. $\endgroup$– Brian M. ScottSep 16, 2011 at 8:27
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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. $\endgroup$– Brian M. ScottSep 16, 2011 at 8:45
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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
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2 Answers
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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.
answered Sep 16, 2011 at 8:16
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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.)
answered Sep 16, 2011 at 8:16