Module 1: Basic Set Theory
Module 2: Modular Arithmetic, Divisibility, and the Fundamental Theorem of Arithmetic
Module 3: Functions and Relations
Module 4: Truth Tables and Symbolic Logic
Module 5: Basic Direct Proofs
Module 6: Proof Techniques Part 1: Contrapositive and Contradiction
Module 7: Sequences, Sums, and Products
Module 8: Proof Techniques Part 2: (Weak) Induction
Module 9: Recurrence Relations and Recursion
Module 10: Counting Systems (Binary, Hex, Octal, etc.)
Module 11: Combinatorics
Module 12: Graph Theory
Module 13: Review

Set Inclusion (Subsets)

In other words, a set \(A\) is a subset of \(B\) if everything that is in \(A\) is also in \(B\), or put another way, \(B\) contains every element that is in \(A\).

We would say that \(A\) is NOT a subset of \(B\) (and write \(A\not\subseteq B\)) if there is at least one element in \(A\) that is NOT in \(B\).

Think of it this way: if \(A\) contains everything that is in \(B\) and \(B\) contains everything that is in \(A\), then they have to contain exactly the same stuff!

Try the following problem!

True!

Every element in \(\{0,1,5\}\) is also an element of \(\{0,1,2,3,4,5\}\). Therefore the former is a subset of the latter.

False!

Notice that \(\{0,1,5\}\) contains the element 0, but the set \(\{1,5,7,11,12\}\) does NOT contain 0. Therefore \(\{0,1,5\}\not\subseteq \{1,5, 7, 11, 12\}\).

True!

The set \(\mathbb{N}\) contains all positive counting numbers, and \(\mathbb{Z}\) contains all positive AND negative whole numbers along with the number 0. So every natural number is indeed an integer (i.e. everything that is in \(\mathbb{N}\) is also in \(\mathbb{Z}\)).

False!

The set \(\{-5, 6, 2,4\}\) has an element that is NOT in the set of natural numbers (i.e. \(\mathbb{N}\)). Which one is it?

True!

\(\{1,2, \{1,2\}\}\) contains three elements; all of which are contained in the set \(\{1,2,3,4,\{1,2\}, \{2,3,9\}\}\). More explicitly, the former set contains 1,2 and the set containing 1 and 2. These three elements appear in the bigger set, making \(\{1,2, \{1,2\}\}\subseteq\{1,2,3,4,\{1,2\}, \{2,3,9\}\}\).

True!

This one is tricky to think about. This question is asking “is everything that is in the empty set (or an empty box) also in the empty set (in another empty box). The answer is “yes.” The first empty box contains nothing, which is also what is in the second empty box. So \(\emptyset\subseteq\emptyset\).

True!

It is helpful to shade both sets on the same number line to answer these sorts of questions. The interval \((1,4)\) contains all numbers between 1 and 4, but NOT 1 and 4, including all decimals and fractions. The interval \([0,5]\) contains all numbers between, and including, 0 through 5. The numbers between 1 and 4 are included in this. So \((1,4)\subseteq [0,5]\)

False!

Note that the interval \((-1,1)\) contains negative numbers (e.g. -0.5, -0.0001, etc.) whereas \([0,2]\) contains all numbers from 0 to 2, including the endpoints. \([0,2]\) contains no negative numbers of any kind. So \((-1,1)\not\subseteq [0,2]\).

True! The interval \((-5,5)\) contains all numbers between -5 and 5, but NOT -5 and 5 themselves. The interval \([-5,5]\) contains all numbers between -5 and 5, AS WELL AS the endpoints -5 and 5. So everything that is in \((-5,5)\) is also in \([-5,5]\).

False! The interval \([0,1]\) contains everything between 0 and 1, AS WELL AS the numbers 0 and 1. The second set contains ALMOST everything that is in the first interval; only one element of \([0,1]\) is not in \((0,1]\). Which is it?

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