
Mathematics for Master Students
Which tasks gave you trouble? Bring them up now: we take the time to go through them before we move on to today’s topic.
Question
Negate: “Some delivery route is unprofitable.”
“Every delivery route is profitable.”: negation swaps \(\exists\) into \(\forall\) and negates the inside.
Common Mistake
“Some route is profitable” is not the negation: both statements can easily be true at the same time.
A set is an unordered collection of distinct objects, called its elements.
Think of a set as a bag: what matters is only what is inside, not in which order it went in, and never twice.
Membership is the most basic set statement:
Every “is it in or not?” question has a clear yes/no answer: that is what makes a set well-defined.
Two standard ways to write a set down:
Set-builder notation is a filter: start from a base set, keep only the elements passing the condition, exactly what a database query does.
The empty set \(\emptyset = \{\}\) contains no elements at all.
Common Mistake
\(\emptyset \neq \{0\}\): the set containing zero has one element. The empty set has none.
\(A \subseteq B\): every element of \(A\) is also in \(B\) (“\(A\) is a subset of \(B\)”).
\(\subseteq\) works like \(\leq\) and \(\subset\) like \(<\): the extra line allows equality.
Two sets are equal if they contain exactly the same elements:
\[A = B \quad \Leftrightarrow \quad A \subseteq B \text{ and } B \subseteq A\]
The classic confusion: \(\in\) relates an element to a set, \(\subseteq\) relates two sets.
Let \(A = \{1, 2, 3\}\):
| Statement | True? | Why |
|---|---|---|
| \(2 \in A\) | ✓ | \(2\) is an element of \(A\) |
| \(\{2\} \subseteq A\) | ✓ | every element of \(\{2\}\) is in \(A\) |
| \(\{2\} \in A\) | ✗ | the set \(\{2\}\) is not one of the three elements |
Common Mistake
Writing \(2 \subseteq A\): a number is not a set, so \(\subseteq\) does not apply. Ask first: element or set?
Question
Let \(B = \{5, 10, 15\}\). Which statements are true: \(\quad 10 \in B, \quad \{5, 15\} \subseteq B, \quad \{10\} \in B, \quad \emptyset \subseteq B\)?
The cardinality \(|A|\) is the number of elements in \(A\):
Common Mistake
Same bars, different meaning: \(|-5| = 5\) is an absolute value (of a number), \(|\{-5\}| = 1\) is a cardinality (of a set). The argument decides.
Session 1 was full of sets, we just did not call them that:
Set operations play out inside a universal set \(U\), all objects currently under discussion:
A Venn diagram shows sets as circles inside the rectangle \(U\):

Overlap = shared elements, outside the circles = the rest of \(U\): a thinking tool, not just a picture.
\(A \cup B = \{x : x \in A \text{ or } x \in B\}\): everything in at least one of the sets.

\(\{1, 2\} \cup \{2, 3\} = \{1, 2, 3\}\): the “or” is inclusive, just like in logic; the shared \(2\) appears once.
\(A \cap B = \{x : x \in A \text{ and } x \in B\}\): only the elements in both sets.

\(\{1, 2\} \cap \{2, 3\} = \{2\}\): think of it as applying two filters at once.
\(A \setminus B = \{x : x \in A \text{ and } x \notin B\}\): what is left of \(A\) after removing everything in \(B\).

\(\{1, 2, 3\} \setminus \{2, 3\} = \{1\}\), and beware: \(A \setminus B \neq B \setminus A\) in general, order matters.
\(A^c = \{x \in U : x \notin A\}\): everything in \(U\) outside of \(A\), i.e. \(A^c = U \setminus A\).

Books also write \(\overline{A}\) or \(A'\): same idea. A handy identity: \(A \setminus B = A \cap B^c\).
\(A\) and \(B\) are disjoint if \(A \cap B = \emptyset\), no shared elements:

Disjoint sets are how we model clean segmentations: every customer is in exactly one region, no overlaps.
Question
For any two sets \(A\) and \(B\): is \(\quad A \cup B \subseteq A \quad\) always true?
Adding \(|A| + |B|\) counts the intersection twice, so subtract it once, the inclusion-exclusion rule:
\[|A \cup B| = |A| + |B| - |A \cap B|\]
Survey example: of 100 coffee drinkers, 70 take sugar, 60 take cream, 50 take both:
\[|A \cup B| = 70 + 60 - 50 = 80\]
So \(100 - 80 = 20\) drink their coffee black: the complement \((A \cup B)^c\).
Complement flips union into intersection, and vice versa:
\[(A \cup B)^c = A^c \cap B^c \qquad\qquad (A \cap B)^c = A^c \cup B^c\]
Business example: \(A\) = delayed shipments, \(B\) = damaged shipments.
“Neither delayed nor damaged” \(= (A \cup B)^c = A^c \cap B^c\) = “on time and intact”.
This is exactly Session 1’s law \(\neg(p \vee q) \Leftrightarrow \neg p \wedge \neg q\), dressed in set notation.
Set operations are logic on membership statements, \(x \in A \cup B\) means \(x \in A \vee x \in B\):
| Logic (Session 1) | Sets (today) |
|---|---|
| \(p \vee q\) (or) | \(A \cup B\) (union) |
| \(p \wedge q\) (and) | \(A \cap B\) (intersection) |
| \(\neg p\) (not) | \(A^c\) (complement) |
| \(p \Rightarrow q\) (implies) | \(A \subseteq B\) (subset) |
If you remember the logic rule, you get the set rule for free, and the other way around.
An ordered pair \((a, b)\) is not a set. Here, order does matter:
Curly braces \(\{\,\}\) = unordered set, round parentheses \((\,)\) = ordered pair. The bracket type carries the meaning.
\(A \times B = \{(a, b) : a \in A, b \in B\}\): all ordered pairs with the first entry from \(A\), the second from \(B\).
Example: \(A = \{1, 2\}\) and \(B = \{x, y\}\):
\[A \times B = \{(1, x), (1, y), (2, x), (2, y)\}\]
Counting: every choice from \(A\) combines with every choice from \(B\), so
\[|A \times B| = |A| \cdot |B|\]
Let \(W = \{1, 2\}\) be warehouses and \(C = \{1, 2, 3\}\) be customers:
The power set \(\mathcal{P}(A)\) is the set of all subsets of \(A\), including \(\emptyset\) and \(A\) itself.
Example: \(A = \{1, 2\}\):
\[\mathcal{P}(A) = \{\emptyset, \{1\}, \{2\}, \{1, 2\}\}\]
Common Mistake
The elements of \(\mathcal{P}(A)\) are themselves sets: \(\{1\} \in \mathcal{P}(A)\), but \(1 \notin \mathcal{P}(A)\).
\[|\mathcal{P}(A)| = 2^{|A|}\]
Question
A carrier may serve any subset of the regions \(\{\text{North}, \text{East}, \text{South}\}\). How many different service portfolios are possible?
\(2^3 = 8\), the elements of \(\mathcal{P}(\{\text{N}, \text{E}, \text{S}\})\):
\[\emptyset, \{\text{N}\}, \{\text{E}\}, \{\text{S}\}, \{\text{N,E}\}, \{\text{N,S}\}, \{\text{E,S}\}, \{\text{N,E,S}\}\]
Note that \(\emptyset\) counts: “serve no region at all” is a valid (if unambitious) portfolio.
Let \(P\) = premium customers, \(C\) = churned customers, \(N\) = newsletter subscribers:
AND/OR/NOT are \(\cap\), \(\cup\), \(^c\)How to read \(\;(P \cap C) \setminus N\;\)? Work inside out:
Common Mistake
When \(\cup\) and \(\cap\) mix, grouping changes the set: \(A \cup (B \cap C) \neq (A \cup B) \cap C\) in general: when in doubt, add parentheses.
You do not need to compute anything to use set notation: most of its value is in making such statements precise.
Every constraint defines a set of allowed values, and constraints combine by intersection:
That’s it for today.
Any Questions?
Draw a Venn diagram whenever a set expression looks confusing: two circles resolve most doubts faster than staring at symbols.
See you there, and bring your questions!
Session 02 - Sets | Dr. Tobias Vlcek | Home