# Mapping: each nonempty finite subset of $\Bbb R$ - sum of its elements

Could you give me a hint how to solve this problem?

Let $D:= \left\{E \subset \mathbb{R} \ | \ 0< \mathrm{card}(E)< + \infty \right\}$. $\phi\colon D\to\mathbb R$ defined by $\phi(E)\sum_{x \in E}x$

Check if $\phi$ is injective or surjective.

Well, I think that it can't be injective because $\mathrm{card}(\mathcal P(\mathbb{R}))>\mathrm{card}(\mathbb{R})$ , so by the pigeonhole principle, there must to at least two subsets with the same sum of elements.

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Simply show lack of injectivity by exhibiting two finite sets with same sum. Simply show surjectivity by exhibiting a special finite set with given sum. Both is very easy. (Think of 1- or 2-elemtn sets) –  Hagen von Eitzen Jan 11 '13 at 11:06
$card(\mathcal P(\mathbb R))$ has nothing to do here. $D$ is a family of finite subsets of $\mathbb R$, which has the same cardinality as $\mathbb R$ –  Adam Jan 11 '13 at 11:09
The title does not match the body. There are sets of real numbers which are not finite. Also, why did yo remove the part where you explain what you think about the problem? –  Asaf Karagila Jan 11 '13 at 14:02
Because you've proved me wrong. It's back now. –  Bilbo Jan 11 '13 at 14:26
Just a comment about notation: you shouldn't write "$\text{card}(E) < \mathord{+}\infty$" because cardinals and $\mathop{\pm}\infty$ are different kinds of objects. There is not generally a sensible way to compare cardinals with points on the extended real number line $[-\infty,+\infty]$. You probably mean to write "$E$ is finite" instead. –  Trevor Wilson Jan 13 '13 at 21:23

Hint: The fact that $\phi$ is not injective follows from the fact that $\{0,x\}$ and $\{x\}$ have the same sum. Surjectivity follows from a similar argument as well.
Observation: $D$ has the same cardinality as $\mathbb R$.
Maybe it's a stupid question, but how come $card(D)=card(\mathbb{R})$ and $\phi$ isn't a bijection? –  Bilbo Jan 11 '13 at 11:56
@Anna: First note that not every function between two sets of the same cardinality is a bijection. Secondly, the hint I gave you should provide you a method of showing that $\phi$ is indeed surjective but is not injective. Therefore it is not a bijection (a bijection is required to be both injective and surjective). –  Asaf Karagila Jan 11 '13 at 12:01