3. Sets, images and inverse images

Learning goals and definitions

Manipulate unions, intersections, complements and Cartesian products; distinguish images from inverse images; prove function properties with their quantifiers. All complements are relative to a fixed universe $U$. Equality $A=B$ means $A\subseteq B$ and $B\subseteq A$.

The union $A\cup B$ contains elements in either set; the intersection $A\cap B$ contains elements in both; $A\setminus B$ contains those in $A$ but not $B$. The power set $\mathcal P(A)$ is the set of all subsets of $A$. The Cartesian product $A\times B$ is the set of ordered pairs $(a,b)$ with $a\in A,b\in B$; in general $A\times B\ne B\times A$.

For a function $f:X\to Y$ and subsets $A\subseteq X$, $B\subseteq Y$, define $f(A)=\{f(x):x\in A\}$ and $f^{-1}(B)=\{x\in X:f(x)\in B\}$. The latter is defined whether or not an inverse function exists.

Theorem: inverse images preserve set operations

For $B,C\subseteq Y$,

$$f^{-1}(B\cup C)=f^{-1}(B)\cup f^{-1}(C),\quad f^{-1}(B\cap C)=f^{-1}(B)\cap f^{-1}(C),$$

and $f^{-1}(Y\setminus B)=X\setminus f^{-1}(B)$.

Proof. For a point $x\in X$, membership in the first left side means $f(x)\in B$ or $f(x)\in C$, exactly membership in the first right side. Replacing “or” by “and” proves the intersection identity. Membership in the complement preimage means $f(x)\notin B$, exactly $x\notin f^{-1}(B)$ within $X$. Each logical equivalence establishes both inclusions at once.

Images preserve unions, but for intersections only $f(A\cap D)\subseteq f(A)\cap f(D)$ is automatic. Indeed an output from an input in both sets lies in both images. In the reverse direction, equal outputs may have arisen from different inputs.

Theorem: injectivity and image intersections

A function $f:X\to Y$ is injective if and only if $f(A\cap D)=f(A)\cap f(D)$ for every $A,D\subseteq X$.

Proof. If $f$ is injective and $y$ lies in both images, write $y=f(a)=f(d)$ with $a\in A,d\in D$. Injectivity gives $a=d\in A\cap D$, yielding the missing inclusion. Conversely, if $f(a)=f(d)$ with $a\ne d$, take $A=\{a\}$ and $D=\{d\}$. Their intersection is empty but the images intersect, contradicting the assumed equality. Thus no distinct inputs share an output.

Worked examples

Noninjectivity is visible in a set identity. For $f(x)=x^2$ on $\mathbb R$, $A=\{-2\}$ and $D=\{2\}$ have disjoint intersection, whereas both images are $\{4\}$. An inverse image behaves differently: $f^{-1}([1,4])=[-2,-1]\cup[1,2]$.

De Morgan with an arbitrary index set. For subsets $A_i\subseteq U$, an element lies outside $\bigcup_iA_i$ exactly when it lies in none of them, equivalently in every complement. Thus $U\setminus\bigcup_iA_i=\bigcap_i(U\setminus A_i)$. For an empty index set, the union is empty and the intersection is $U$, so the convention remains consistent.

Graded practice

1. Foundation. Prove $A\setminus B=A\cap B^c$ by membership.
2. Core. For $f:\mathbb R\to\mathbb R$, $f(x)=x^2-1$, find $f([-2,1])$ and $f^{-1}((-1,3])$.
3. Core. Prove $A\subseteq f^{-1}(f(A))$. When does equality hold for every subset $A$?
4. Proof. Prove $f(f^{-1}(B))=B\cap f(X)$ and deduce a characterization of surjectivity.

Full solutions

1. For $x\in U$, $x\in A\setminus B$ means $x\in A$ and $x\notin B$, exactly $x\in A\cap B^c$.
2. On $[-2,1]$, $x^2$ takes every value in $[0,4]$, so the image is $[-1,3]$. The preimage condition is $-1<x^2-1\le3$, or $0<x^2\le4$, giving $[-2,0)\cup(0,2]$.
3. If $x\in A$, then $f(x)\in f(A)$, proving the inclusion. For injective $f$, any $x$ in the right side has $f(x)=f(a)$ with $a\in A$, hence $x=a\in A$. Conversely equality for singleton $A=\{a\}$ rules out a distinct input with the same output. Thus the condition is exactly injectivity.
4. Membership in the left side means $y=f(x)$ for some $x$ with $f(x)\in B$, exactly $y\in B$ and $y\in f(X)$. Therefore equality to $B$ for every $B\subseteq Y$ holds exactly when $f(X)=Y$, which is surjectivity (take $B=Y$ for the reverse implication).

Summary and term definitions

An image follows outputs of a chosen input set; an inverse image collects all inputs whose outputs enter a chosen target set. A power set consists of subsets as its elements. Double inclusion proves set equality in two directions. Writing membership conditions prevents invalid cancellation of a function and exposes exactly where injectivity or surjectivity is needed.