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According to Definition 2.2 in Baby Rudin, the preimage of a set E, denoted $f^{-1}(E)$, is the set of all $x$ such that $f(x) \in E$.

Suppose that $f(x)=x^2$, the set A is the interval [0,1], and the set B is the interval [-1,1].

Then it would appear to me that $f^{-1}(A)=[-1,1]$, and $f^{-1}(B)=\{ [i,0),[0,1] \}$, since $f([i,0])$=[-1,0].

Is this correct? It seems that an interval is typically by definition a set of real numbers. Is it then allowed to introduce complex numbers in this way?

Asaf Karagila
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ormondo
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  • One would assume all these sets are subsets of the reals. Of course you did not specify that, but it is likely understood in these exercises. At any rate if you try to include complex numbers, working with inequalities does not work, no order relation on the complex numbers. – coffeemath Jun 26 '23 at 09:07
  • You need to be clear about the domain and codomain where your map is defined. If your function is $f:[-2,2]\to \mathbb{R}$ by $f(x)=x^2$, then $f^{-1}([0,1]) = [-1,1]$ and $f^{-1}([-1,1]) = {x\in [-2,2] : f(x)\in [-1,1]}=[-1,1]$. The preimage (or inverse image) is always a subset of the domain of the function. So if the domain is a subset of real numbers, the preimage of any subset of the codomain will be a real numbers subset too. – Psaro Jun 26 '23 at 09:09
  • Thank you both. Yes, the exercise doesn't say, but I suppose the domain of $f$ is the real numbers, as you suggest. @Psaro, What's odd to me is this: How can there be any $x \in [−2,2]$ such that $f(x) \in [−1,0]$? Don't we get a positive real number for all $x$ in that range? Or is it perhaps a mistake on my part to assume that there is an $x$ such that $f(x)∈[−1,0]$? – ormondo Jun 26 '23 at 10:07
  • If you change the domain or codomain you change the function. So you should expect the properties of the function to change if you adjust the domain or codomain. – CyclotomicField Jun 26 '23 at 11:07
  • I think some of the comments here are a little misleading, since not all definitions of function include the concept of a "codomain". In particular, it is common in both analysis and set theory to identify a function with its graph, meaning that the "codomain" of a function is ill-defined. There is more discussion of this here. – Joe Jun 26 '23 at 11:19
  • @ormondo You are right, you are assuming there is $x\in[-2,2]$ such that $f(x)\in[-1,0]$. And that's not true. – Psaro Jun 26 '23 at 12:22

1 Answers1

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If $E$ is any set, and if $f:X\to Y$ is a function, then we can make sense of $f^{-1}(E)$: it is the set of elements $x\in X$ such that $f(x)\in E$.*

In your case, you haven't specified the domain of $f$, but without further context, I would assume that it is $\mathbb R$. In that case, $$ f^{-1}([-1,1])=\{x\in\mathbb R:x^2\in[-1,1]\} \, . $$ The above set equals $[-1,1].$ We can prove this as follows: we know that $[-1,1]\subset f^{-1}([-1,1])$, since if $x\in [-1,1]$, then $x^2\in[-1,1].$ Moreover, $f^{-1}([-1,1]\subset[-1,1]$, since if $x\in\mathbb R$ and $x\not\in[-1,1]$, then $x^2>1$, so $x\not\in f^{-1}([-1,1]).$

The fact that there is no $x\in\mathbb R$ such that $x^2\in[-1,0]$ is not relevant here. To drive this point home, consider $f^{-1}(\{i,-i\})$: it is the set of $x\in\mathbb R$ such that $x^2\in\{i,-i\}.$ This set makes perfect sense, but of course there is no $x\in\mathbb R$ such that $x^2\in\{-i,i\}$, and so $f^{-1}(\{i,-i\})=\varnothing.$

If you intended the domain of $f$ to be $\mathbb C$, then $f^{-1}([-1,1])=[0,1]\cup\{ix:x\in[0,1]\}$. It's worth noting that notation like $[i,0)$ is non-standard: we typically only use interval notation for subsets of $\mathbb R$. Moreover, $\{a,b\}$ is a set with two elements $a$ and $b$ (unless $a=b$, in which case the set only has one element).

*Occasionally, this notation is ambiguous – when $f$ is invertible and $E$ is an element of the range of $f$, and a subset of the range of $f$, then it is unclear what $f^{-1}(E)$ means: does it refer to the set of $x\in X$ such that $f(x)\in E$, or does it mean the unique $y\in X$ such that $f(y)=E$? For example, if $f:\{42\}\to\{\varnothing\}$ is given by $f(42)=\varnothing$, then $\varnothing$ is a subset of the range of $f$, and an element of the range of $f$. In practice, these issues rarely arise.

Joe
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