Binary operations on various sets defined by set-builder notation

Question

Assume:

• functions $f:X\rightarrow Y$ and $g:X\rightarrow Y$
• predicates (boolean-valued functions) $P:X \rightarrow \{\mathrm{true}, \mathrm{false}\}$ and $Q:X \rightarrow \{\mathrm{true}, \mathrm{false}\}$
• subsets $A,A_1,A_2 \subseteq X$.

Are the following statements true? And what is a simple proof in each case (or a reference that would discuss them)?

1. One function, two subsets:

• $\{f(x):x\in A_1\} \cup \{f(x):x\in A_2\} = \{f(x):x\in A_1 \cup A_2\}$ ?
• $\{f(x):x\in A_1\} \cap \{f(x):x\in A_2\} = \{f(x):x\in A_1 \cap A_2\}$ ?
• $\{f(x):x\in A_1\} \setminus \{f(x):x\in A_2\} = \{f(x):x\in A_1 \setminus A_2\}$ ?
• $\{f(x):x\in A_1\} \triangle \{f(x):x\in A_2\} = \{f(x):x\in A_1 \triangle A_2\}$ ?

2. Two functions, one subset:

• $\{f(x):x\in A\} \cup \{g(x):x\in A\} =$ (any nontrivial form) ?
• $\{f(x):x\in A\} \cap \{g(x):x\in A\} =$ (any nontrivial form) ?
• $\{f(x):x\in A\} \setminus \{g(x):x\in A\} =$ (any nontrivial form) ?
• $\{f(x):x\in A\} \triangle \{g(x):x\in A\} =$ (any nontrivial form) ?

3. One predicate, two subsets:

• $\{x\in A_1 : P(x)\} \cup \{x\in A_2:P(x)\} = \{x\in A_1 \cup A_2:P(x)\}$ ?
• $\{x\in A_1:P(x)\} \cap \{x\in A_2:P(x)\} = \{x\in A_1 \cap A_2:P(x)\}$ ?
• $\{x\in A_1:P(x)\} \setminus \{x\in A_2:P(x)\} = \{x\in A_1 \setminus A_2:P(x)\}$ ?
• $\{x\in A_1:P(x)\} \triangle \{x\in A_2:P(x)\} = \{x\in A_1 \triangle A_2:P(x)\}$ ?

4. Two predicates, one subset:

• $\{x\in A:P(x)\} \cup \{x\in A:Q(x)\} = \{x\in A:P(x) \lor Q(x)\}$ ?
• $\{x\in A:P(x)\} \cap \{x\in A:Q(x)\} = \{x\in A:P(x) \land Q(x)\}$ ?
• $\{x\in A:P(x)\} \setminus \{x\in A:Q(x)\} = \{x\in A:P(x) \nrightarrow Q(x)\}$ ?
• $\{x\in A:P(x)\} \triangle \{x\in A:Q(x)\} = \{x\in A:P(x) \oplus Q(x)\}$ ?

Background

Originally, I was trying to decide whether, for function $f:X\rightarrow Y$ and subsets $A_1,A_2 \subseteq X$, the following is true:

$$\{f(x):x\in A_1\} \cup \{f(x):x\in A_2\} = \{f(x):x\in A_2 \cup A_2\}\,?$$

Then I started wondering about various related questions, including:

• extending this to other set operations ($\cup, \cap,\setminus,\triangle$)
• using just one subset $A \subseteq X$ but different functions $f:X \rightarrow Y$ and $g:X \rightarrow Y$
• using set-builder forms based on predicates, like $\{x \in A:P(x)\}$, instead of replacement, like $\{f(x):x \in A\}$ (a.k.a. "filters" and "maps" in functional programming).

For the sake of completeness, I have listed all combinations of these variations in the question above. Some of them intuitively seem true, but I'm having trouble convincing myself, and I'm having trouble searching for proofs/discussions of these particular situations, both on this site and in textbooks.

Answer 1

As to the original question, the answer is yes: $f(A\cup B) = f(A)\cup f(B)$.

Let us prove the inclusion $(\subseteq)$ first.

If $y\in f(A\cup B)$, then $y = f(x)$ and $x\in A\cup B$. If $x\in A$, then $y\in f(A)\subseteq f(A)\cup f(B)$.

If $x\in B$, then $y\in f(B)\subseteq f(A)\cup f(B)$.

In both cases, $y\in f(A)\cup f(B)$, and we are done.

Let us prove the inclusion $(\supseteq)$ now.

If $y\in f(A)\cup f(B)$, then $y\in f(A)$ or $y\in f(B)$.

If $y\in f(A)$, then $y = f(x)$ and $x\in A\subseteq A\cup B$. Then $y\in f(A\cup B)$.

If $y\in f(B)$, then $y = f(x)$ and $x\in B\subseteq A\cup B$. Then $y\in f(A\cup B)$.

In both cases, $y\in f(A\cup B)$, and we are done.

Comment

As to the other questions, it is advisable to ask one question at a time.

Hopefully this helps.

Answer 2

Short Answer

1. No, not equal in general, but equal if $f$ is injective.
2. Nothing useful.
3. Yes.
4. Yes.

(Thanks to user1337 for the suggestion to express things in terms of images.)

Definitions

For any function $f:X \rightarrow Y$ and subset $A \subseteq X$, the image of $A$ under $f$ is:

$$f(A) = \{f(x):x \in A\}.$$

For subset $B \subseteq Y$, the preimage of $B$ under $f$ is:

$$f^{-1}(B) = \{x \in X:f(x) \in B\}.$$

There are 16 possible binary operations (which correspond to the 16 logical connectives). Here we partition these into two groups of 8, which we call Type 1 and Type 2. For sets $I$ and $J$, there are 8 Type 1 binary operations $*$ such that $I * J \subseteq I \cup J$ always holds. These are:

$$\varnothing \quad I\!\setminus\!J \quad I\!\cap\!J \quad I \quad J\!\setminus\!I \quad I \triangle J \quad J \quad I\!\cup\!J,$$

and there are 8 Type 2 binary operations $*$ (which are just the complements of Type 1) where $I * J \subseteq I \cup J$ does not necessarily hold:

$$\varnothing^c \quad (I\!\setminus\!J)^c \quad (I\!\cap\!J)^c \quad I^c \quad (J\!\setminus\!I)^c \quad (I \triangle J)^c \quad J^c \quad (I\!\cup\!J)^c.$$

Note that some of these are also nullary or unary operations, but they can be treated as binary operations which simply ignore one or both arguments.

Part 1: One function/image, two subsets

For this case, the answer is found in many resources which discuss the image (and preimage) and its many basic properties, e.g. whether it preserves set operations. Most textbooks on set theory or topology discuss these, usually in introductory chapters, appendices, or as proof exercises. For example:

• Overview of basic results about images and preimages
• Wikipedia: Image (mathematics)
• ProofWiki pages on mappings and relations (especially "Image of ..." pages)
• p.388 of Lee, 2010, Introduction to Topological Manifolds, 2nd Ed.
• pp.19-21 of Munkres, 2000, Topology, 2nd Ed.
• p.39 of Halmos, 1960, Naive Set Theory.

In particular, for function $f:X \rightarrow Y$ and subsets $A_1,A_2 \subseteq X$, the following properties hold in general:

• $f(A_1) \cup f(A_2) = f(A_1 \cup A_2)$
• $f(A_1) \cap f(A_2) \supseteq f(A_1 \cap A_2)$
• $f(A_1) \setminus f(A_2) \subseteq f(A_1 \setminus A_2)$
• $f(A_1) \triangle f(A_2) \subseteq f(A_1 \triangle A_2)$,

and equality holds in all four cases if $f$ is injective.

Generalization to All Binary Operations

For the 8 Type 1 operations $*$, for sets $I,J$, we have the following relations $R$ for $f(I) \! * \! f(J) \; R \; f(I\!*\!J)$:

operation $*$	$R$ (general or surjective $f$)	$R$ (injective or bijective $f$)
$\varnothing$	$=$	=
$I\!\setminus\!J$	$\subseteq$	=
$I\!\cap\!J$	$\supseteq$	=
$I$	$=$	=
$J\!\setminus\!I$	$\subseteq$	=
$I \triangle J$	$\subseteq$	=
$J$	$=$	=
$I\!\cup\!J$	$=$	=

And for the 8 Type 2 operations $*$, we have the following relations $R$ (where NR means no relation):

operation $*$	$R$ (general $f$)	$R$ (injective $f$)	$R$ (surjective $f$)	$R$ (bijective $f$)
$\varnothing^c$	$\supseteq$	$\supseteq$	$=$	$=$
$(I\!\setminus\!J)^c$	NR	$\supseteq$	$\subseteq$	$=$
$(I\!\cap\!J)^c$	NR	$\supseteq$	$\subseteq$	$=$
$I^c$	NR	$\supseteq$	$\subseteq$	$=$
$(J\!\setminus\!I)^c$	NR	$\supseteq$	$\subseteq$	$=$
$(I \triangle J)^c$	NR	$\supseteq$	NR	$=$
$J^c$	NR	$\supseteq$	$\subseteq$	$=$
$(I\!\cup\!J)^c$	NR	$\supseteq$	$\subseteq$	$=$

To prove these 16 cases, I used the following visual proof technique:

• Make a Venn diagram involving the four sets $f(I\!\setminus\!J)$, $f(J\!\setminus\!I)$, $f(I\!\cap\!J)$, $f((I\!\cup\!J)^c)$. This diagram thus has 16 regions.
• For each operation $*$, mark the set of regions representing $f(I)\!*\!f(J)$ with one symbol and those representing $f(I\!*\!J)$ with another symbol. (Use the fact that $f(I)\!\cup\!f(J) = f(I\!\cup\!J)$.)
• For injective $f$, mark all 11 "overlapping" regions as empty.
• For surjective $f$, mark the one "outside" region as empty.
• Recall that bijective $f$ is both injective and surjective.

The result makes it easy to see what type of relation exists between the two sets, if any. I used this technique to generate the tables above. Please comment if you notice any errors here.

Part 2: Two functions/images, one subset

For functions $f:X \rightarrow Y$ and $g:X \rightarrow Y$, subset $A \subseteq X$, and binary set operation $* \in \{\cup, \cap, \setminus, \triangle\}$, these images can be written more concisely as:

$$\{f(x) : x \in A\} * \{g(x) : x \in A\} = f(A) * g(A).$$

Probably nothing general can be said in this case without further assumptions.

Part 3: One predicate, two subsets

Define predicate $P:X \rightarrow \{\mathrm{true}, \mathrm{false}\}$. For brevity, let $T=\{\mathrm{true}\}$ and $F=\{\mathrm{false}\}$.

We can express sets defined by a predicate as a preimage:

\begin{aligned} \{x \in X : P(x)\} &= \{x \in X : P(x) \in T\}\\ &= P^{-1}(T), \end{aligned}

i.e. the preimage of $T$ under $P$. We can also focus on any subset $A \subseteq X$:

\begin{aligned} \{x \in A : P(x)\} &= \{x \in A : P(x) \in T\}\\ &= A \cap P^{-1}(T). \end{aligned}

Thus, for subsets $A_1,A_2 \subseteq X$ and binary set operation $* \in \{\cup, \cap, \setminus, \triangle\}$ we have:

\begin{aligned} \{x \in A_1 : P(x)\} * \{x \in A_2 : P(x)\} &= (A_1 \cap P^{-1}(T)) * (A_2 \cap P^{-1}(T))\\ &= (A_1 * A_2) \cap P^{-1}(T)\\ &= \{x \in A_1 * A_2 : P(x)\}. \end{aligned}

This confirms the hypothesis in the original question above.

Generalization to All Binary Operations

We can extend this result to all 16 binary set operations as follows.

For all 8 Type 1 operations $*$, intersection $\cap$ is (left- and right-) distributive over $*$, so for sets $I,J,K$, we have $(I * J) \cap K = (I \cap K) * (J \cap K)$, a fact which is used in the derivation above. Thus, for these 8 operations, the above result applies, i.e.:

\begin{aligned} \{x \in A_1 : P(x)\} * \{x \in A_2 : P(x)\} &= (A_1 * A_2) \cap P^{-1}(T)\\ &= \{x \in A_1 * A_2 : P(x)\}. \end{aligned}

For all 8 Type 2 operations $*$, we instead have $\big((I * J) \cap K\big) \cup K^c = (I \cap K) * (J \cap K)$. Thus, the result becomes:

\begin{aligned} \{x \in A_1 : P(x)\} * \{x \in A_2 : P(x)\} &= (A_1 \cap P^{-1}(T)) * (A_2 \cap P^{-1}(T))\\ &= \big((A_1 * A_2) \cap P^{-1}(T)\big) \cup (P^{-1}(T))^c\\ &= (A_1 * A_2) \cup P^{-1}(F)\\ &= \{x \in A_1 * A_2 : P(x)\} \cup P^{-1}(F)\\ &\supseteq \{x \in A_1 * A_2 : P(x)\}. \end{aligned}

Part 4: Two predicates, one subset

(This case is fairly obvious by inspection, but for completeness, we'll follow a similar derivation as in part 3 above.)

Define predicates $P:X \rightarrow \{\mathrm{true}, \mathrm{false}\}$ and $Q:X \rightarrow \{\mathrm{true}, \mathrm{false}\}$ and subset $A \subseteq X$. Again, let $T=\{\mathrm{true}\}$ and $F=\{\mathrm{false}\}$.

Expressing sets as preimages (cf. part 3 above):

\begin{aligned} \{x \in A : P(x)\} &= A \cap P^{-1}(T)\\ \{x \in A : Q(x)\} &= A \cap Q^{-1}(T). \end{aligned}

For binary set operation $* \in \{\cup, \cap, \setminus, \triangle\}$ with corresponding logical connective $\star \in \{\lor, \land, \nrightarrow, \oplus\}$:

\begin{aligned} \{x \in A : P(x)\} * \{x \in A : Q(x)\} &= (A \cap P^{-1}(T)) * (A \cap Q^{-1}(T))\\ &= A \cap (P^{-1}(T) * Q^{-1}(T))\\ &= A \cap (\{x \in X : P(x)\} * \{x \in X : Q(x)\})\\ &= A \cap \{x \in X : P(x) \star Q(x)\}\\ &= \{x \in A : P(x) \star Q(x)\}. \end{aligned}

This confirms the hypothesis in the original question above.

Generalization to All Binary Operations

We can extend this result to all 16 binary set operations in a similar way as in part 3 above.

For all 8 Type 1 operations $*$, the above result still holds:

\begin{aligned} \{x \in A : P(x)\} * \{x \in A : Q(x)\} &= A \cap \{x \in X : P(x) \star Q(x)\}\\ &= \{x \in A : P(x) \star Q(x)\}. \end{aligned}

For all 8 Type 2 operations $*$, we have:

\begin{aligned} \{x \in A : P(x)\} * \{x \in A : Q(x)\} &= (A \cap P^{-1}(T)) * (A \cap Q^{-1}(T))\\ &= A^c \cup \Big(A \cap \big(P^{-1}(T) * Q^{-1}(T)\big)\Big)\\ &= (A^c \cup A) \cap (A^c \cup \{x \in X : P(x) \star Q(x)\})\\ &= A^c \cup \{x \in X : P(x) \star Q(x)\}\\ &= \{x \in X : (x \notin A) \lor (P(x) \star Q(x))\}\\ &= \{x \in A : P(x) \star Q(x)\} \cup A^c\\ &\supseteq \{x \in A : P(x) \star Q(x)\}. \end{aligned}

Special case: One Preimage, Two Subsets

Let $A=X$, let $B_1,B_2 \subseteq Y$, and let $f:X \rightarrow Y$. Assume predicates take the form $P(x)=(f(x) \in B_1)$ and $Q(x)=(f(x) \in B_2)$ for all $x \in X$. Then we are simply dealing with preimages of $B_1$ and $B_2$ under $f$. Then, using the results above for Type 1 operations $*$:

\begin{aligned} f^{-1}(B_1) * f^{-1}(B_2) &= \{x \in X : P(x)\} * \{x \in X : Q(x)\}\\ &= \{x \in X : P(x) \star Q(x)\}\\ &= \{x \in X : (f(x) \in B_1) \star (f(x) \in B_2)\}\\ &= \{x \in X : f(x) \in B_1 * B_2\}\\ &= f^{-1}(B_1 * B_2). \end{aligned}

For Type 2 operations $*$, we get the same result:

\begin{aligned} f^{-1}(B_1) * f^{-1}(B_2) &= X^c \cup \{x \in X : P(x) \star Q(x)\}\\ &= \varnothing \cup \{x \in X : (f(x) \in B_1) \star (f(x) \in B_2)\}\\ &= \{x \in X : f(x) \in B_1 * B_2\}\\ &= f^{-1}(B_1 * B_2). \end{aligned}