Membership and Containment Are Different Claims
Set arguments often turn on keeping track of what kind of object is being discussed. The statement \(x\in A\) says that the object \(x\) is an element of the set \(A\). The statement \(A\subseteq B\) says that every element of \(A\) is also an element of \(B\). These statements have different meanings, and neither can replace the other without justification. The basic proof method is element chasing: start with an arbitrary element of the set in question, and use the definitions to show where it belongs.
We use the standard meanings of set-builder notation and membership: \(\{x\in X:P(x)\}\) is the set of elements \(x\) of \(X\) for which the condition \(P(x)\) holds, and \(x\in A\) means that \(x\) is an element of \(A\). The empty set, written \(\varnothing\), is the set with no elements.
The quantifier “every” matters. To prove \(A\subseteq B\), it is not enough to find an element shared by the two sets. One must take an arbitrary \(x\in A\) and show that this \(x\) lies in \(B\). Conversely, a single element \(x\in A\) with \(x\notin B\) disproves \(A\subseteq B\). If \(A=\varnothing\), the subset condition is true for every \(B\): there is no element of \(A\) that could fail to belong to \(B\).
Worked Example: Membership Is Not Containment
Let \(A=\{1,4,7\}\) and \(B=\{n\in\mathbb{Z}:1\le n\le 8\}\). Decide whether \(4\in A\), whether \(A\subseteq B\), and whether \(C\subseteq B\) for \(C=\{1,4,9\}\).
First, \(4\in A\), because \(4\) is one of the three listed elements of \(A\). This is a membership statement about a single object.
To check \(A\subseteq B\), take an arbitrary \(x\in A\). The definition of \(A\) gives three possibilities: \(x=1\), \(x=4\), or \(x=7\). Each is an integer between \(1\) and \(8\), inclusive: \(1\le1\le8\), \(1\le4\le8\), and \(1\le7\le8\). Thus each possible \(x\) belongs to \(B\), so \(A\subseteq B\).
For \(C\subseteq B\), the element \(9\) is a counterexample: \(9\in C\), but \(9\notin B\), since \(9\le8\) is false. Therefore \(C\not\subseteq B\).
Check: The verification of \(A\subseteq B\) considered every element of \(A\), whereas the failure of \(C\subseteq B\) required only the one element \(9\). The fact that \(1\in C\) and \(1\in B\) does not settle whether \(C\subseteq B\).
How to Prove a Subset Claim
The definition gives a reliable proof pattern. Begin with “Let \(x\in A\) be arbitrary.” Then use the information \(x\in A\) to obtain a property of \(x\), and show that this property implies \(x\in B\). The word “arbitrary” is essential: the reasoning must not depend on choosing a special element that happens to work.
Assume \(x\in A\), without making an extra assumption about which element of \(A\) it is.
Use the description of \(A\) to identify what must be true of \(x\).
Show that the property of \(x\) implies \(x\in B\). This proves \(A\subseteq B\).
Proof. Assume \(A\subseteq B\) and \(B\subseteq C\). To prove \(A\subseteq C\), take an arbitrary \(x\in A\). Since \(A\subseteq B\), the definition of subset gives \(x\in B\). Since \(B\subseteq C\), it then gives \(x\in C\). We have shown that every element of \(A\) is an element of \(C\), so \(A\subseteq C\). If \(A=\varnothing\), there is no \(x\in A\) to consider, and the defining condition for \(A\subseteq C\) holds vacuously. \(\square\)
Worked Example: A Subset Claim With a Boundary Case
In \(\mathbb{R}\) with its usual number order, let \(A=\{x\in\mathbb{R}:0\le x<2\}\) and \(B=\{x\in\mathbb{R}:x^2<4\}\). Prove that \(A\subseteq B\), and determine whether \(B\subseteq A\).
Take an arbitrary \(x\in A\). Then \(0\le x<2\). We must show \(x^2<4\). If \(x=0\), then \(x^2=0<4\). If \(x>0\), multiplying \(x<2\) by the positive number \(x\) gives \(x^2<2x\). Also, multiplying \(x<2\) by the positive number \(2\) gives \(2x<4\). Therefore \(x^2<2x<4\), so \(x^2<4\). In either case \(x\in B\). Since \(x\) was arbitrary, \(A\subseteq B\).
For the reverse inclusion, test \(x=-1\). We have \((-1)^2=1<4\), so \(-1\in B\). But \(0\le-1\) is false, so \(-1\notin A\). Therefore \(B\not\subseteq A\), and in particular \(A\ne B\).
Check: The calculation for \(x=0\) gives \(x^2=0<4\) directly. The multiplication step \(x<2\Rightarrow x^2<2x\) was used only when \(x>0\), so it does not incorrectly assert a strict inequality after multiplying by zero.
Equality Requires Two Inclusions
Equality says that the elements of the two sets match exactly. To verify it, prove that every element of \(A\) lies in \(B\), and then prove that every element of \(B\) lies in \(A\). These are separate tasks: one direction establishes only that \(A\) has no elements outside \(B\), while the other rules out elements of \(B\) outside \(A\).
Proof. Suppose first that \(A=B\). If \(x\in A\), then \(x\) is an element of \(B\), because the sets have exactly the same elements. Thus \(A\subseteq B\). Reversing the roles of \(A\) and \(B\) gives \(B\subseteq A\).
Conversely, suppose \(A\subseteq B\) and \(B\subseteq A\). If \(x\in A\), the first inclusion gives \(x\in B\). If \(x\in B\), the second inclusion gives \(x\in A\). Hence an object belongs to \(A\) exactly when it belongs to \(B\), so the sets have exactly the same elements and \(A=B\). This reasoning also covers an empty set: if one set has no elements, the two inclusions require the other to have no elements either. \(\square\)
Worked Example: Equality of Two Solution Sets
In \(\mathbb{R}\), prove that \(A=\{x\in\mathbb{R}:x^2=4\}\) equals \(B=\{-2,2\}\).
First prove \(A\subseteq B\). Let \(x\in A\). Then \(x^2=4\), so \((x-2)(x+2)=x^2-4=0\). A product of two real numbers is zero only if at least one factor is zero. Thus \(x-2=0\) or \(x+2=0\), giving \(x=2\) or \(x=-2\). In either case \(x\in B\).
Now prove \(B\subseteq A\). Let \(x\in B\). By the definition of \(B\), either \(x=-2\) or \(x=2\). If \(x=-2\), then \(x^2=(-2)(-2)=4\). If \(x=2\), then \(x^2=2\cdot2=4\). Thus \(x\in A\) in both cases. The double-inclusion criterion now gives \(A=B\).
Check: Substituting the two listed elements into the defining equation gives \((-2)^2=4\) and \(2^2=4\). The factorization argument also shows that no other real number can belong to \(A\).
A Second Equality Proof: Divisibility
Element chasing works just as well when set membership is described by an integer parameter. In that situation, the proof must show that the parameter required on one side can be found on the other side.
Worked Example: Even Integers in Two Descriptions
Let \(A=\{n\in\mathbb{Z}: n=2k\text{ for some }k\in\mathbb{Z}\}\) and \(B=\{n\in\mathbb{Z}: n=2m\text{ for some }m\in\mathbb{Z}\}\). Prove \(A=B\).
Take an arbitrary \(n\in A\). By the definition of \(A\), there is an integer \(k\) such that \(n=2k\). Taking \(m=k\), which is an integer, gives \(n=2m\). Hence \(n\in B\), and \(A\subseteq B\).
For the reverse direction, take an arbitrary \(n\in B\). There is an integer \(m\) such that \(n=2m\). Taking \(k=m\) gives \(n=2k\), so \(n\in A\). Hence \(B\subseteq A\), and the double-inclusion criterion yields \(A=B\).
Check: The argument includes \(n=0\): choose \(k=0\) and \(m=0\), and \(0=2\cdot0\) in both descriptions. No nonzero condition was assumed.
Common Mistakes and a Practical Check
Common mistake: confusing one successful element with a subset proof. For \(A=\{2,5\}\) and \(B=\{2,3\}\), the calculation \(2\in A\) and \(2\in B\) does not prove \(A\subseteq B\). The attempted argument checks only one element. The correction is to test every element required by the definition: \(5\in A\), but \(5\notin B\), so \(A\not\subseteq B\).
Common mistake: treating one inclusion as equality. In the first worked example, \(A=\{1,4,7\}\) is contained in \(B=\{n\in\mathbb{Z}:1\le n\le8\}\), but \(A\ne B\), since \(2\in B\) and \(2\notin A\). The correction is to prove both inclusions, or, when disproving equality, to exhibit an element that belongs to one set and not the other.
Before finishing a proof, check which quantifier the claim requires. A membership claim concerns a particular element. A subset claim must work for every element of the proposed smaller set. An equality claim requires both subset claims. When the membership condition uses inequalities, also check boundary cases: multiplying a strict inequality by a number that might be zero can turn a strict inequality into equality, as the \(x=0\) case above illustrates.
Check Your Understanding
For each question, state the membership or inclusion claim precisely before deciding whether it is true.
- Let \(A=\{0,2,4\}\) and \(B=\{n\in\mathbb{Z}:0\le n\le4\}\). Is \(2\in A\)? Is \(A\subseteq B\)? Explain the difference between the two answers.
- Let \(C=\{x\in\mathbb{R}:x^2=9\}\) and \(D=\{-3,3\}\). Prove \(C=D\) by double inclusion.
- Give a specific element that shows \(\{1,3,6\}\not\subseteq\{1,3,5\}\).
- Prove from the definition that \(\varnothing\subseteq A\) for every set \(A\).
- For \(E=\{x\in\mathbb{R}:0\le x\le1\}\) and \(F=\{x\in\mathbb{R}:x^2\le1\}\), decide whether \(E\subseteq F\). In your proof, check \(x=0\) separately if you use multiplication by \(x\).