a Subset Phenomenon
Taxonomy in Biology to Identify the Ladder of Life
In biology, the science of classifying all living things is called taxonomy. More than 2000 years ago, Aristotle formalized animal classification with his “ladder of life”: higher animals, lower animals, higher plants. lower piants. To- day, living organisms are classified into six kingdoms (or sets) called animalia, plantae, archaea, eubacteria, fungi, and protista. Even more general groupings of living things are made according to shared characteristics. The groupings. from most general to most specific, are kingdom, phylum, ciass, order, family, genus. and species. For example, a zebra. Equus burchelli, is a member of the genus Equus, as is the horse, Equus caballus. Both the zebra and the horse are mem- bers of the universal set called the kingdom of animals and the same family, Equidae; they are members of different species (E. burchelli and E. caballus), however.
Why) This is Important
Scientists use sets to classify and categorize animals, plants, and all forms of life.
Definition: Subset
Set A is a subset of set B, symbolized by A⊆B, if and only if all the elements of set A are also elements of set B.
The symbol A⊆B indicates that “set A is a subset of set B.” The symbol ⊈ is used to indicate “is not a subset.” Thus, A⊈B indicates that set A is not a subset of set B.
To show that set A is not a subset of set B, we must find at least one element of set A that is not an elentent of set B.
⛲ Example 1: A Subset?
Determine whether set N is a subset of set O.
a) N={rain, snow, sleet} O={rain, snow, sleet, hail}
b) N={q,r,s,t} O={q,r}
c) N={x|x is a yellow fruit} O={x|x is a red fruit}
d) N={vanilla, chocolate, rocky road} O={chocolate, vanilla, rocky road}
✍ Solution:
a) All the elements of set N are contained in set O, so N⊆O.
b) The elements s and t are in set N but not in set O. so N⊈O (N is not a subset of O). In this example. however. all the elements of set O are contained in set N; therefore. O⊆N.
c) There are fruits, such as bananas, that are in set N that are not in set O, so N⊈O.
d) All the elements of set N are contained in set O, so N⊆O. Note also that O⊆N. In fact, set N =set O.
Definition: Proper Subset
Set A is a proper subset of set B. symbolized by A⊂B, if and only if all the elements of set A are elements of set B and set A≠ set B (that is, set B must contain at least one element not in set A).
Consider the sets G={red. blue, yellow} and H={red. orange. yellow, green, blue. violet}. Set G is a subset of set H, G⊆H, because every element of set G is also an element of set H. Set G is also a proper subset of set H. G⊂H, because set G and set H are not equal. Now consider J={ear. bus, train} and K={train, car, bus}. Set J is a subset of set K, C⊆D, because every element of set J is also an element of set K. Set J, however, is not a proper subset of set K, J⊄K. because set J and set K are equal sets.
Proper Subset
If L⊆M and L≠M, then L is called a proper subset of M, written as L⊂M and M is called proper superset of L.
e.g., Let L={x:x is an even natural number}
and M={x:x is a natural number}.
Then, L={2,4,6,8,…}
and M={1,2,3,4,5,…}
∴
⛲ Example 2: A Proper Subset?
Determine whether set P is a proper subset of set Q.
a) P={jazz, pop, hip hop} Q={classical, jazz, pop, rap, hip hop}
b) P={a,b,c,d} Q={a,c,b,d}
✍ Solution:
a) All the elements of set P are contained in set Q. and sets Pand Q are not equal; thus, P⊂Q.
b) Set P=set Q, so P⊄Q. (However, P⊆Q.)
Every set is a subset of itself, but no set is a proper subset of itself. For all sets P, P⊆P, but P⊄P. For example, if P={1,2,3} then P⊆P because every element of set P is contained in set P , but P⊄P because set P =set P.
⛲ Ex3. Consider the following sets Ø, R={1,2} and S={1,4,8}.
Insert the following symbols ⊂ or ⊄ between each of the following pair of sets.
(i) Ø … S (ii) R … S
✍ Solution:
(i) Since, null set is proper subset of every set. ∴
(ii) Given, R={1,2} and S={1,4,8}. Since, element 2∉S. ∴
⛲ Ex4. Consider the sets
Insert the symbol ⊂ or ⊄ between each of the following pair of sets:
(i) Ø … U (ii) T … U (iii) T … V (iv) U … V
✍ Solution:
(i) Ø⊂U as Ø is a subset of every set.
(ii) A⊄U as 3∈T and 3∉U
(iii) T⊂V as 1,3∈T also belongs to V
(iv) U⊂V as each element of U is also an element of V.
⛲ Ex5. Decide,among the following sets,which sets are subsets of one and another: W={x:x∈ℝ and x satisfy x2-8x+12=0}, X={2,4,6}, Y={2,4,6,8,…}, Z={6}.
✍ Solution:
W={x:x∈ℝ and x satisfy x2-8x+12=0}
2 and 6 are the only solutions of x2-8x+12=0
∴W={2,6}
X={2,4,6}, Y={2,4,6,8,…}, Z={6}
∴
Hence, W⊂X,W⊂Y,X⊂Y,Z⊂W,Z⊂X,Z⊂Y.
⛲ Ex6. Make correct statements by filling in the symbols ⊂ or ⊄ in the blank spaces:
(i) {2,3,4} … {1,2,3,4,5}
(ii) {a,b,c} … {b,c,d}
(iii) {x:x is a student of Class XI of your school} … {x:x student of your school}
(iv) {x:x is a circle in the plane} … {x:x is a circle in the same plane with radius 1 unit}
(v) {x:x is a triangle in a plane} … {x:x is a rectangle in the plane}
(vi) {x:x is an equilateral triangle in a plane} … {x:x is a triangle in the same plane}
(vii) {x:x is an even natural number} … {x:x is an integer}
✍
(i) {2,3,4}⊂{1,2,3,4,5}
(ii) {a,b,c}⊄{b,c,d}
(iii) {x:x is a student of class XI of your school}⊂{x:x is student of your school}
(iv) {x:x is a circle in the plane}⊄{x:x is a circle in the same plane with radius 1 unit}
(v) {x:x is a triangle in a plane}⊄{x:x is a rectangle in the plane}
(vi) {x:x is an equilateral triangle in a plane}⊂{x:x in a triangle in the same plane} (vii) {x:x is an even natural number}⊂{x:x is an integer}
🌈 Element ∊ or Proper Subset ⊂ — True or False Statements
