It it well-defined to map to a set? That is, $$f(x):= \{x,x+1\}$$ for example? If so, how would one define it? $f: \mathbb{R} \rightarrow ...$ what? What does one need to be wary of in order to ensure well-definedness? What about injectivity/subjectivity issues?
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At the moment, you don't map a function to a set, you define a function which maps numbers to sets, which is allowed. Then $f$ goes to $\left{ { x } \mid x \in \mathbb{R} \right}$. – Hetebrij Sep 29 '15 at 12:20
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In settheory "everything is a set". Also the elements of e.g. $\mathbb R$. In that context a function sends sets to sets. In your example you could define $\wp_2(\mathbb R)$ as the collection of subsets of $\mathbb R$ that have $2$ elements. Then it can serve as codomain for the function. There are more possibilities. Also $\wp(\mathbb R)$ itself will do. – drhab Sep 29 '15 at 12:25
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You can use $\mathcal{P}(\mathbb{R})$, which means the "set of subsets of $\mathbb{R}$". Then you can for sure define a function from $\mathbb{R}$ to $\mathcal{P}(\mathbb{R})$ by saying that $f(x) = \{x,x+1\}$. For injectivity or surjectivity there is no difficulties... Your map is injective since two different real numbers are mapped on two different sets. It is not surjective since the set $\{1,2,3\}$ is not in the image, for example.
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The sole requirement for an object to be a function on a set $X$ to a set $Y$ is that it assigns to every $x \in X$ some unique $y \in Y$; we can assign to each $x \in \mathbb{R}$ the unique set $\{x, x+1 \} \in 2^{\mathbb{R}}$, so we legitimately obtain the function $x \mapsto \{ x, x+1\}: \mathbb{R} \to 2^{\mathbb{R}}$.
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