Images of Unions and Intersections

What Happens to Unions and Intersections Under a Function?

A function sends each point of its domain to one point of its codomain. When we apply a function to a set, different domain points may have the same image. This possibility explains an important difference between unions and intersections: images always preserve unions, but images need not preserve intersections. We will prove the general statements, identify when intersection equality holds, and keep track of what changes for an empty family.

Let $f:X\to Y$ be a function, and let $A\subseteq X$. The image of $A$ under $f$ is the subset of $Y$ consisting of the values $f(x)$ for $x\in A$. The image is a subset of the codomain $Y$, even if it does not contain every point of $Y$.

Definition: For $A\subseteq X$, the image of $A$ is $f(A)=\{f(x):x\in A\}\subseteq Y$. For an indexed family $(A_i)_{i\in I}$ of subsets of $X$, we use $\bigcup_{i\in\varnothing}A_i=\varnothing$ and $\bigcap_{i\in\varnothing}A_i=X$. An empty intersection is taken in the ambient set containing its members; thus, for subsets of $Y$, $\bigcap_{i\in\varnothing}f(A_i)=Y$.

Images Preserve Arbitrary Unions

Theorem 1: Let $f:X\to Y$ be a function and let $(A_i)_{i\in I}$ be any indexed family of subsets of $X$, with $I$ possibly empty. Then $$ f\left(\bigcup_{i\in I}A_i\right)=\bigcup_{i\in I}f(A_i). $$

Proof. Let $y\in Y$. By the definition of image, $y\in f(\bigcup_{i\in I}A_i)$ if and only if there is some $x\in\bigcup_{i\in I}A_i$ such that $f(x)=y$. Membership in the union means that there is an index $i\in I$ with $x\in A_i$. Thus the condition is equivalent to there being an $i\in I$ such that $y\in f(A_i)$, which is equivalent to $y\in\bigcup_{i\in I}f(A_i)$. This proves equality by membership in both sets. If $I=\varnothing$, both unions are empty, and $f(\varnothing)=\varnothing$ by the definition of image. $\square$

The argument allows any number of sets in the family. The reason it works is that membership in a union already asserts the existence of an index and a point in the corresponding set. Applying $f$ carries that same point into the corresponding image.

Worked Example: An Infinite Union Under a Constant Function

Let $X=\mathbb{Z}$, $Y=\{p\}$, and $f:\mathbb{Z}\to\{p\}$ be defined by $f(k)=p$ for every $k\in\mathbb{Z}$. For each $n\in\mathbb{N}$, set $A_n=\{n\}$. Compare the image of the union with the union of the images.

First, $\bigcup_{n\in\mathbb{N}}A_n=\mathbb{N}$, since each $A_n$ contains exactly $n$. As $\mathbb{N}$ is nonempty and $f(n)=p$ for every $n$, its image is $f(\mathbb{N})=\{p\}$.

For each individual index $n$, $f(A_n)=f(\{n\})=\{f(n)\}=\{p\}$. Therefore $\bigcup_{n\in\mathbb{N}}f(A_n)=\bigcup_{n\in\mathbb{N}}\{p\}=\{p\}$. The two sides are both $\{p\}$, as Theorem 1 predicts.

Check: $f(\bigcup_n A_n)=f(\mathbb{N})=\{p\}$, and $\bigcup_n f(A_n)=\bigcup_n\{p\}=\{p\}$. This calculation uses the same value $p$ at each index; it does not require the function to be injective.

Images of Intersections: Inclusion Always Holds

Theorem 2: Let $f:X\to Y$ be a function and $(A_i)_{i\in I}$ any indexed family of subsets of $X$, including an empty family. Then $$ f\left(\bigcap_{i\in I}A_i\right)\subseteq\bigcap_{i\in I}f(A_i). $$

Proof. First suppose $I\ne\varnothing$. Let $y\in f(\bigcap_{i\in I}A_i)$. By the definition of image, there is an $x\in\bigcap_{i\in I}A_i$ with $f(x)=y$. Since $x\in A_i$ for every $i\in I$, we have $y=f(x)\in f(A_i)$ for every $i\in I$. Hence $y\in\bigcap_{i\in I}f(A_i)$, proving the inclusion. If $I=\varnothing$, the left side is $f(X)$, while the right side is $Y$. Since $f:X\to Y$, every value $f(x)$ lies in $Y$, so $f(X)\subseteq Y$. Thus the inclusion holds in this case as well. $\square$

The inclusion can be strict. A point in every $f(A_i)$ may come from different points of the different sets $A_i$. To conclude that it comes from one point lying in every $A_i$, we need a condition that rules out distinct points having the same image.

Definition: A function $f:X\to Y$ is injective if, whenever $x,x'\in X$ and $f(x)=f(x')$, it follows that $x=x'$.

Theorem 3: If $f:X\to Y$ is injective and $I\ne\varnothing$, then for every family $(A_i)_{i\in I}$ of subsets of $X$, $$ f\left(\bigcap_{i\in I}A_i\right)=\bigcap_{i\in I}f(A_i). $$

Proof. The inclusion from left to right is Theorem 2. For the reverse inclusion, choose one index $i_0\in I$, which is possible because $I\ne\varnothing$, and let $y\in\bigcap_{i\in I}f(A_i)$. In particular, $y\in f(A_{i_0})$, so there is an $x_0\in A_{i_0}$ such that $f(x_0)=y$. For each $i\in I$, membership $y\in f(A_i)$ gives a point $x_i\in A_i$ with $f(x_i)=y$. Injectivity implies $x_i=x_0$. Therefore $x_0\in A_i$ for every $i$, so $x_0\in\bigcap_{i\in I}A_i$. Since $f(x_0)=y$, we have $y\in f(\bigcap_{i\in I}A_i)$. This proves the reverse inclusion and hence equality. $\square$

The proof uses a single index to get one point $x_0$; injectivity then forces every other point with the same image to equal $x_0$. It does not require choosing one point independently from every set in the family. The nonempty-index condition matters: for an empty family, the claimed equality becomes $f(X)=Y$, which does not follow from injectivity.

Worked Example: Equality for an Injective Function

Let $f:\mathbb{Z}\to\mathbb{Z}$ be $f(k)=2k+1$, and take $A_1=\{0,1,2\}$ and $A_2=\{1,2,4\}$. Compute both sides of the intersection identity.

The common elements of the two domain sets are $1$ and $2$, so $A_1\cap A_2=\{1,2\}$. Applying $f$ gives $f(A_1\cap A_2)=\{2(1)+1,2(2)+1\}=\{3,5\}$.

For the other side, $f(A_1)=\{2(0)+1,2(1)+1,2(2)+1\}=\{1,3,5\}$, while $f(A_2)=\{2(1)+1,2(2)+1,2(4)+1\}=\{3,5,9\}$. Their intersection is $\{3,5\}$. Thus both sides equal $\{3,5\}$. The function is injective: if $2k+1=2\ell+1$, subtracting $1$ and dividing by $2$ gives $k=\ell$, so Theorem 3 applies.

Check: The common domain elements $1,2$ map to $3,5$, and these are exactly the common values in $\{1,3,5\}$ and $\{3,5,9\}$.

Worked Example: Strict Inclusion for a Noninjective Function

Let $f:\mathbb{R}\to\mathbb{R}$ be $f(x)=x^2$, with $A_1=\{-2,1\}$ and $A_2=\{2,3\}$. Compare the image of the intersection with the intersection of the images.

The sets $A_1$ and $A_2$ have no common elements, so $A_1\cap A_2=\varnothing$, and $f(A_1\cap A_2)=f(\varnothing)=\varnothing$. On the other hand, $f(A_1)=\{(-2)^2,1^2\}=\{4,1\}=\{1,4\}$, and $f(A_2)=\{2^2,3^2\}=\{4,9\}$. Consequently $f(A_1)\cap f(A_2)=\{4\}$.

Therefore $\varnothing\subsetneq\{4\}$. The value $4$ is in both image sets, but it comes from $-2\in A_1$ and $2\in A_2$; those are different points, and neither belongs to $A_1\cap A_2$. The function is not injective because $f(-2)=4=f(2)$ while $-2\ne2$.

Check: The only common value in $\{1,4\}$ and $\{4,9\}$ is $4$, whereas the image of the empty intersection contains no values. This verifies the strict inclusion directly.

Empty Families and a Common Pitfall

The conditions on the intersection statements are worth recording precisely. The inclusion from Theorem 2 needs no nonempty-family assumption. For an empty family it says $f(X)\subseteq Y$, which is true for every function $f:X\to Y$. In contrast, the equality in Theorem 3 is stated for a nonempty family. If the family is empty, equality holds exactly when $f(X)=Y$; this is the condition that every element of $Y$ is a value of $f$.

Worked Example: An Empty Intersection

Let $X=\{a\}$, $Y=\{0,1\}$, and define $f(a)=0$. This function is injective: the only possible pair of domain points with equal images is $a,a$, and those points are equal. Consider the empty family of subsets of $X$.

By the empty-intersection convention, its intersection in $X$ is $X=\{a\}$, so the image of that intersection is $f(X)=\{0\}$. The intersection of the empty family of subsets of $Y$ is $Y=\{0,1\}$. Thus $f(X)=\{0\}\subseteq\{0,1\}=Y$, but the two sets are not equal.

Check: $f(X)$ contains the value $0$ and does not contain $1$, while the empty intersection on the codomain side is all of $Y$, including $1$. This shows why injectivity alone does not give equality for an empty family.

Common pitfall. One might claim $f(A\cap B)=f(A)\cap f(B)$ for every function because the inclusion always holds. The square-function calculation above disproves that claim: its left side is $\varnothing$, while its right side is $\{4\}$. The correction is to keep the direction of the general statement as an inclusion and use injectivity to obtain equality for nonempty families.

Operation	General statement	Condition for equality
Union	$f(\bigcup_{i\in I}A_i)=\bigcup_{i\in I}f(A_i)$	No additional condition, including when $I$ is empty
Intersection	$f(\bigcap_{i\in I}A_i)\subseteq\bigcap_{i\in I}f(A_i)$	No additional condition is needed for the inclusion
Intersection, $I\ne\varnothing$	$f(\bigcap_{i\in I}A_i)=\bigcap_{i\in I}f(A_i)$	Holds when $f$ is injective
Intersection, $I=\varnothing$	$f(X)\subseteq Y$	Equality holds exactly when $f(X)=Y$

The practical distinction is simple: union membership supplies a point in at least one member of the family, so taking images preserves the union exactly. For intersections, a common image value might arise from different domain points. Injectivity prevents that ambiguity when the family is nonempty; the empty-family case instead compares the whole image $f(X)$ with the whole codomain $Y$.

Check Your Understanding

For each question, state the ambient set for an empty intersection and justify your answer from the definitions.

Let $f:\mathbb{Z}\to\mathbb{Z}$ be $f(n)=n+1$. For $A_1=\{0,1\}$ and $A_2=\{1,2\}$, compute $f(A_1\cap A_2)$ and $f(A_1)\cap f(A_2)$.
Give a function and two subsets for which $f(A\cap B)$ is a proper subset of $f(A)\cap f(B)$.
Why does $f(\bigcup_{i\in I}A_i)=\bigcup_{i\in I}f(A_i)$ remain true when $I=\varnothing$?
For a nonempty family $(A_i)_{i\in I}$, identify the step in the intersection-equality proof that uses injectivity.
For the empty family, what condition on $f:X\to Y$ makes $f(\bigcap_{i\in\varnothing}A_i)=\bigcap_{i\in\varnothing}f(A_i)$ true?

Operation	General statement	Condition for equality
Union	\(f(\bigcup_{i\in I}A_i)=\bigcup_{i\in I}f(A_i)\)	No additional condition, including when \(I\) is empty
Intersection	\(f(\bigcap_{i\in I}A_i)\subseteq\bigcap_{i\in I}f(A_i)\)	No additional condition is needed for the inclusion
Intersection, \(I\ne\varnothing\)	\(f(\bigcap_{i\in I}A_i)=\bigcap_{i\in I}f(A_i)\)	Holds when \(f\) is injective
Intersection, \(I=\varnothing\)	\(f(X)\subseteq Y\)	Equality holds exactly when \(f(X)=Y\)