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I have $\mathfrak P(\mathbb{R})$ being the set of all subsets of $\mathbb{R}$, meaning $\mathfrak P(\mathbb{R}) = \{X|X\subseteq \mathbb{R}\}$. I then have $F$ being the set of all functions $\mathbb{R} \rightarrow \mathbb{R}$. Define $Z:F \rightarrow \mathfrak P(\mathbb{R}) $ by $Z(f) = \{x \in \mathbb{R}|f(x) = 0 \}$ and I am trying to work out if Z is injective or surjective.

I feel as though mapping all functions, onto an infinite number of infinite sized subsets would surely be injective and not surjective. How can I start a proof of this?${}$

I still don't get it.

Katie
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2 Answers2

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You need to think about what the map $Z$ means. Saying that $Z$ is surjective means that for any subset of $\mathbb{R}$, there is some map $\mathbb{R}\to\mathbb{R}$ that has that subset as its set of zeros. Saying that $Z$ is injective says that if two functions from $\mathbb{R}$ to $\mathbb{R}$ have the same set of zeros, then they are equal. If you understand those statements, you should easily be able to answer the question.

rogerl
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I'll furnish the details to rogerl's answer.

Surjectivity: Note that $Z$ can either be surjective or not surjective. To show that it is surjective one needs to show that for ANY subset $X\subseteq \mathbb{R}$, we have a function, $f$ say, such that $f(x)=0$ for all $x\in X$.

The function $f(x)=\begin{cases} 0&\text{if } x\in X\\ 1&\text{if } x\notin X\\ \end{cases}$ springs to mind.

Injectivity: If $Z$ is injective, then if $f,g:\mathbb{R}\rightarrow\mathbb{R}$ have the same set of zeros, then it must be that $f=g$. Consider $f(x)=x$, $g(x)=x^2$, they have the same set of zero, namely $0$ itself, but they are not the same function, hence, $Z$ is not injective.

BlackAdder
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