Sets and functions: domain logic before algebra

Learning goals and role

Use set notation accurately, distinguish a simultaneous condition from an alternative branch, and determine domains before manipulating formulas.

1. Elements, subsets and the empty set

The statement $x\in A$ says $x$ is an element. The statement $A\subseteq B$ says every element of $A$ is also in $B$. They have different logical types. Sets are determined by their members, not by order or repetition: $\{1,2,2\}=\{2,1\}$. The empty set $\varnothing$ has no elements and is a subset of every set, since there is no element that violates the inclusion.

The set $\{\varnothing\}$ is not empty: it contains one element, namely the empty set. A power set $\mathcal P(A)$ is the set of all subsets of $A$; it always contains $\varnothing$. This does not mean $\varnothing\in A$ for every $A$. For example the real interval $[0,1]$ contains numbers, not a set serving as an extra endpoint.

For real inputs, interval notation records endpoint membership: $(a,b)$ excludes both endpoints, $[a,b]$ includes them, and a half-open interval includes just the indicated square-bracket endpoint. Infinity is not a real member, so an interval extending to infinity always uses a parenthesis there.

2. Operations translate logical connectors

The complement depends on a universal set $E$. In this unit we usually assume all sets live inside some fixed $E$, so $A^c$ means $E \setminus A$.

Write $U$ for the fixed universe $E$ above. In set-builder notation,

$$A\cup B=\{x\in U:x\in A\text{ or }x\in B\},\quad A\cap B=\{x\in U:x\in A\text{ and }x\in B\},$$ $$A^c=U\setminus A,\qquad A\setminus B=\{x\in A:x\notin B\}.$$

The word “or” is inclusive. Complements depend on the universe, while $A\setminus B$ is determined by the two named sets.

Proposition: De Morgan and difference rules

For subsets of the fixed universe, $(A\cup B)^c=A^c\cap B^c$, $(A\cap B)^c=A^c\cup B^c$, and $A\setminus B=A\cap B^c$.

Proof. An element fails to belong to a union exactly when it belongs to neither set, proving the first rule by equivalent membership statements. It fails to belong to an intersection exactly when at least one membership fails, proving the second. Membership in the difference means membership in $A$ together with nonmembership in $B$, the third rule. Because every step is an equivalence, both set inclusions have been proved, not only one.

For finite sets, $|A\cup B|=|A|+|B|-|A\cap B|$: the first addition counts the intersection twice, and subtracting one copy corrects it. A Cartesian product $A\times B$ contains ordered pairs $(a,b)$, so its finite cardinality is $|A||B|$: for each of the $|A|$ first entries there are $|B|$ second entries. This is a counting statement, not an assumption that probabilistic events are independent.

3. A function needs one output at every declared input

Definition

A function

A function from $X$ to $Y$ is a subset of $X \times Y$ such that every $x \in X$ is paired with exactly one $y \in Y$.

For $f:X\to Y$, $X$ is the domain and $Y$ the codomain. Its image or range is $f(X)$ and need not fill $Y$. The graph is a subset of $X\times Y$ containing exactly one ordered pair above every input. A formula with an undefined value at an allegedly included input does not define a function on that declared domain.

For a target subset $B\subseteq Y$, the inverse image $f^{-1}(B)=\{x\in X:f(x)\in B\}$ is defined even if the function has no inverse function. An inverse function on the full codomain instead requires bijectivity: injectivity prevents two inputs sharing an output, and surjectivity ensures every target has one.

Worked domain choice. The rule $x\mapsto x^2$ from $\mathbb R$ to $[0,\infty)$ is surjective but not injective. Restricting its domain to $[0,\infty)$ makes it bijective, with inverse $\sqrt y$. Declaring its codomain to be all $\mathbb R$ would make it non-surjective even after that restriction. Domain and codomain are mathematical data, not labels added after solving an equation.

4. Domain intersections for simultaneous operations

For a sum, product or quotient of two real functions, an input must belong to both original domains. A quotient also requires a nonzero denominator. A square root requires a nonnegative radicand, and a logarithm a strictly positive argument. These are simultaneous conditions and therefore intersect.

Let $f_a(x)=\sqrt{a-x}/(x+1)$ with real parameter $a$. Its domain is $\{x:x\le a\}\setminus\{-1\}$. To make the dependence visible:

• If $a<-1$, the domain is $(-\infty,a]$; the excluded value is already outside it.
• If $a=-1$, it is $(-\infty,-1)$; the radicand would allow the endpoint but the denominator excludes it.
• If $a>-1$, it is $(-\infty,-1)\cup(-1,a]$.

Now form $g_a(x)=\ln(f_a(x))$. The logarithm requires the quotient to be positive. Its square-root numerator cannot be negative, so positivity requires $x<a$ and $x>-1$. Thus the domain is $(-1,a)$ when $a>-1$, and empty when $a\le-1$. Taking logs changes the domain; it is not enough to keep the radicand and denominator conditions alone. An empty domain is a legitimate mathematical possibility, though it may show a proposed real model has no permissible inputs.

5. Piecewise alternatives produce a union

Consider the rule using $\sqrt{-x}$ when $x\le0$ and $\ln(x-1)$ when $x>0$. The first branch is valid on $(-\infty,0]$. The second branch is valid only when both $x>0$ and $x-1>0$, giving $(1,\infty)$. The actual domain is the union $(-\infty,0]\cup(1,\infty)$.

It is not their intersection: no input is required to use both alternative branches. It is also not all real numbers: the branch instruction $x>0$ does not magically make $\ln(x-1)$ defined on $(0,1]$. Piecewise specifications need valid formulas on their assigned branches and compatible outputs wherever two branch domains overlap. If two overlapping branches give different outputs, the rule fails the single-output function requirement.

For comparison, the ordinary sum $\sqrt{-x}+\ln(x-1)$ would require both formulas simultaneously, so its real domain is empty. The same two formulas lead to different domain logic when combined as alternatives rather than added.

6. Algebraic equations still need the original input set

The equation $\sqrt{x+2}=x$ forces $x\ge0$ because the left side is nonnegative. Squaring gives $x^2=x+2$, whose roots are two and minus one. Only two satisfies the original equation. The squared equation generated candidates; the original sign condition selected the actual solution.

Likewise cancellation in $(x^2-1)/(x-1)$ gives $x+1$ only on the original domain $x\ne1$. The formula $x+1$ on all reals is a different function, an extension with a filled-in missing point. Equal algebraic values on a common domain do not identify functions with different domains.

Graded set-language and model tasks

1. Membership. Explain why $\varnothing\subseteq A$ always holds but $\varnothing\in A$ does not. Compute $\mathcal P(\varnothing)$ and $\mathcal P(\{0\})$.
2. Universe. For $A=[0,1]$, compare its complement in $U=\mathbb R$ with its complement in $V=[-1,2]$. Prove why the answers differ without changing $A$.
3. Domain logic. Find the real domain of $\ln(x^2-1)/\sqrt{4-x^2}$, identifying why each endpoint or zero is included or excluded.
4. Piecewise function. A rule gives $x$ when $x\le1$ and $2-x$ when $x\ge0$. Does it define a function on all reals as written? Locate every conflict and explain how to repair the specification without claiming a unique repair.
5. Parameter composition. For $g_a(x)=\ln(\sqrt{a-x}/(x+1))$, prove that its domain is nonempty exactly when $a>-1$, and explain why the two strict endpoints arise from different conditions.
6. Function identity. Are $f(x)=(x^2-1)/(x-1)$ on its natural domain and $h(x)=x+1$ on all reals equal functions? State the precise equality relation that does hold and how to define an extension.

Full solutions

1. A failed inclusion would require an element of the empty set outside $A$, impossible. Membership is a separate statement about whether $A$ contains that set as one of its elements, false for example at $A=\{0\}$. The only subset of the empty set is itself, so $\mathcal P(\varnothing)=\{\varnothing\}$. The subsets of $\{0\}$ are $\varnothing$ and $\{0\}$, giving a two-element power set.
2. In $\mathbb R$ the complement is $(-\infty,0)\cup(1,\infty)$. In $[-1,2]$ it is $[-1,0)\cup(1,2]$. A complement consists of universe members not in $A$, so the second removes the outside-universe portions while retaining the same nonmembership test. Complements must always specify their ambient set.
3. The logarithm requires $x^2-1>0$, or $x<-1$ or $x>1$. The denominator requires a defined nonzero square root, so $4-x^2>0$, or $-2<x<2$. Their intersection is $(-2,-1)\cup(1,2)$. The values $\pm1$ make the log argument zero; $\pm2$ make the denominator zero; all four are excluded for different reasons.
4. Both branches apply on $[0,1]$. Their outputs agree only when $x=2-x$, namely $x=1$. They conflict at every $x\in[0,1)$. One repair is to use the first branch for $x<1$ and the second for $x\ge1$; another can choose a different nonoverlapping split. The original conditions do not select a unique repair, so silently imposing one would change the definition.
5. The square root must be strictly positive to be a positive logarithm input, giving $x<a$. Its denominator must be positive because the numerator cannot be negative, giving $x>-1$. Thus admissible inputs exist exactly when the strict interval $(-1,a)$ is nonempty. The right endpoint makes the numerator zero, and the left makes the quotient denominator zero; both are invalid for different stages of the composition.
6. Their domains differ, so they are not equal as functions. For every $x\ne1$, factorization gives $f(x)=h(x)$, meaning $f$ equals the restriction of $h$ to $\mathbb R\setminus\{1\}$. Defining a new function to agree with $f$ there and assigning value two at one produces exactly the extension $h$. The original function does not acquire that value by cancellation alone.

Summary and terms

A set operation encodes a logical connector. A domain intersection collects simultaneous requirements, while a piecewise-domain union collects valid alternatives. A restriction reduces a function's domain; an extension adds inputs with assigned compatible outputs. An inverse image is defined for every function, unlike an inverse function. Correct set language makes formula domains, branch conditions and model validity explicit before algebra begins.