2.2: Comparing Sets
- Page ID
- 52923
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Two sets A and B are equal if they have exactly the same elements. We write A = B.
Two sets A and B are equivalent if \(n(A)\)= \(n(B)\). Another way of saying this is that two sets are equivalent if they have the same number of elements.
Determine if the following pairs of sets are equal, equivalent, or both
- {1, 3, 5, 7, 9} and {7, 5, 1, 3, 9}
- \(\emptyset\) and {x : x is a living human born before 1600}
- {Facebook, Instagram, Tik Tok, Snapchat} and {baseball, hockey, football, basketball}
Solution
- These two sets are equal because they have the same exact elements. It does not matter that they are rearranged. They are also equivalent because they both have 5 elements.
- These two sets are equal because there are no humans alive that were born before 1600. They are also equivalent because they both have 0 elements.
- These two sets are not equal because one set is social media platforms and the other set is sports. They are, however, equivalent because they both have 4 elements.
Sometimes a collection might not contain all the elements of a set. For example, Chris owns three Madonna albums. While Chris’s collection is a set, we can also say it is a subset of the larger set of all Madonna albums.
A subset of a set \(A\) is another set that contains only elements from the set \(A\), but may not contain all the elements of \(A\).
If \(B\) is a subset of \(A,\) we write \(B \subseteq A\)
A proper subset is a subset that is not identical to the original set – it contains fewer elements.
If \(B\) is a proper subset of \(A\), we write \(B \subset A\)
Consider these three sets
\(A=\) the set of all even numbers\(\quad B=\{2,4,6\} \quad C=\{2,3,4,6\}\)
Here \(B \subset A\) since every element of \(B\) is also an even number, so is an element of \(A\).
More formally, we could say \(B \subset A\) since if \(x \in B,\) then \(x \in A\)
It is also true that \(B \subset C\).
\(C\) is not a subset of \(A\), since \(C\) contains an element, 3 , that is not contained in \(A\)
Suppose a set contains the plays “Much Ado About Nothing”, “MacBeth”, and “A Midsummer’s Night Dream”. What is a larger set this might be a subset of?
Solution
There are many possible answers here. One would be the set of plays by Shakespeare. This is also a subset of the set of all plays ever written. It is also a subset of all British literature.
The set \(A=\{1,3,5\} .\) What is a larger set this might be a subset of?
- Answer
-
There are several answers: The set of all odd numbers less than 10. The set of all odd numbers. The set of all integers. The set of all real numbers.
One way to build intuition about subsets is to try listing all the different subsets of a particular set. Let's look at some examples of small sets and identify all of their subsets.
List all of the subsets for the following sets:
a. The empty set \(\emptyset\)
b. {a}
c. {m, n}
d. {x, y, z}
Solution
a. Since the empty set has zero elements, the only subset of the empty set \(\emptyset\) is the empty set itself.
b. The set {a} has two subsets: the set {a} itself as well as the empty set \(\emptyset\)
c. The set {m, n} has four subsets: the empty set \(\emptyset\), {m}, {n} and {m, n}
d. The set {x, y, z} has eight subsets: the empty set \(\emptyset\), {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, and {x, y, z}
There are several things to observe here. First notice that every set has itself as a subset. Also, the empty set is a subset of every possible set. Lastly, did you see what happened as the sets increased in size? For each new element of the set - 0, 1, 2, 3 - the number of subsets doubles - 1, 2, 4, 8. This pattern continues for sets of any size, so we can come up with a formula to predict the number of subsets for a given set.
A set with k elements has 2k different subsets.