Session 02 - Sets

Mathematics for Master Students

Dr. Tobias Vlcek

Recap & Your Questions

Session 01 in a Nutshell

  • Number sets form a hierarchy: \(\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}\)
  • Intervals like \([0, 500]\) describe connected ranges of reals
  • \(\sum\) and \(\prod\) are compact loops for adding and multiplying
  • Indices carry meaning: \(x_{ij}\) = flow from warehouse \(i\) to customer \(j\)
  • Implication \(p \Rightarrow q\) is a one-way street: only the contrapositive is equivalent
  • Negation swaps quantifiers: \(\neg\,(\forall x: P(x)) \Leftrightarrow \exists x: \neg P(x)\)

Common Issues in the Homework

  • Counting terms: \(\sum_{i=3}^{10} x_i\) has \(10 - 3 + 1 = 8\) terms, not \(7\): both endpoints count!
  • Over-strong negation: negating “some warehouse operates above \(90\%\) capacity” gives “every warehouse runs at at most \(90\%\)”, not “every warehouse is idle”
  • Converse error: from “certified \(\Rightarrow\) passed audit” and a supplier that passed the audit, nothing follows about its certification

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.

Warm-Up: One More Negation

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.

Today’s Plan

  • Set basics: elements, subsets, and cardinality
  • Set operations: union, intersection, difference, complement
  • Products & power sets: ordered pairs, routes, and selections
  • Sets in business: segmentation and feasible sets

Set Basics

What Is a Set?

A set is an unordered collection of distinct objects, called its elements.

  • Transport modes: \(M = \{\text{road}, \text{rail}, \text{waterway}, \text{ocean}, \text{air}\}\)
  • Order does not matter: \(\{1, 2, 3\} = \{3, 1, 2\}\)
  • No duplicates: \(\{1, 2, 2, 3\} = \{1, 2, 3\}\)

Think of a set as a bag: what matters is only what is inside, not in which order it went in, and never twice.

Element Notation

Membership is the most basic set statement:

  • \(a \in A\): “\(a\) is an element of \(A\)”
  • \(a \notin A\): “\(a\) is not an element of \(A\)”
  • \(\text{rail} \in M\), but \(\text{drone} \notin M\)
  • From Session 1: \(3 \in \mathbb{N}\), while \(-3 \notin \mathbb{N}\) (but \(-3 \in \mathbb{Z}\))

Every “is it in or not?” question has a clear yes/no answer: that is what makes a set well-defined.

Describing Sets

Two standard ways to write a set down:

  • Listing the elements: \(A = \{2, 4, 6, 8\}\)
  • Set-builder notation: \(A = \{x \in \mathbb{N} : x \text{ is even and } x \leq 8\}\)
  • Read “\(:\)” as such that: some books write “\(\mid\)” instead
  • For infinite sets, listing fails, set-builder still works: \(\{x \in \mathbb{N} : x \text{ is even}\}\)

Set-builder notation is a filter: start from a base set, keep only the elements passing the condition, exactly what a database query does.

Set-Builder in Practice

The same filter pattern appears throughout your courses:

  • Key accounts: \(\{c \in \text{Customers} : \text{revenue}(c) > 10000\}\)
  • Late shipments: \(\{s \in \text{Shipments} : \text{arrival}(s) > \text{promised}(s)\}\)
  • Feasible order sizes: \(\{q \in \mathbb{N}_0 : 20 \leq q \leq 500\}\)

Every set-builder expression has the same three parts: a base set, a colon, and a condition. Change the condition, and you have a new filter.

Your Turn: How Many Pass the Filter?

Question

How many elements does \(\;\{x \in \mathbb{Z} : x^2 \leq 4\}\;\) have? Shout the number.

Five: the set is \(\{-2, -1, 0, 1, 2\}\).

Common Mistake

Answering \(3\) (only \(0, 1, 2\)) forgets that the base set is \(\mathbb{Z}\): negative integers square to positive numbers, so \(-1\) and \(-2\) pass the filter too.

The Empty Set

The empty set \(\emptyset = \{\}\) contains no elements at all.

  • The set of all shipments that arrived before they were sent: \(\emptyset\)
  • \(\emptyset\) is a subset of every set: it has no element that could be missing
  • An empty result is an answer, not an error: “no customer matches the filter”

Common Mistake

\(\emptyset \neq \{0\}\): the set containing zero has one element. The empty set has none.

Subsets

\(A \subseteq B\): every element of \(A\) is also in \(B\) (“\(A\) is a subset of \(B\)”).

  • \(\{2, 3\} \subseteq \{1, 2, 3\}\), and also \(\{1, 2, 3\} \subseteq \{1, 2, 3\}\)
  • Proper subset \(A \subset B\): \(A \subseteq B\) and \(A \neq B\)
  • Session 1’s hierarchy \(\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}\) is a chain of proper subsets
  • For every set: \(A \subseteq A\) and \(\emptyset \subseteq A\)

\(\subseteq\) works like \(\leq\) and \(\subset\) like \(<\): the extra line allows equality.

Checking a Subset Claim

Let \(F = \{x \in \mathbb{N} : x \text{ is divisible by } 4\}\) and \(E = \{x \in \mathbb{N} : x \text{ is even}\}\).

  • \(F \subseteq E\)? Take any element of \(F\): a multiple of \(4\) is \(4k = 2 \cdot 2k\), hence even. Yes
  • \(E \subseteq F\)? Look for a counterexample: \(2 \in E\), but \(2 \notin F\). No
  • So \(F \subset E\) is a proper subset: every multiple of \(4\) is even, but not the other way round

To prove \(A \subseteq B\), argue for an arbitrary element. To disprove it, one counterexample suffices: Session 1’s \(\forall\) versus \(\exists\) again.

Set Equality

Two sets are equal if they contain exactly the same elements:

\[A = B \quad \Leftrightarrow \quad A \subseteq B \text{ and } B \subseteq A\]

  • \(\{1, 2, 3\} = \{3, 2, 1\}\): same bag, different listing order
  • To check equality, check both inclusions: a pattern you will see in many courses

Element vs Subset

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?

Your Turn: Element or Subset?

Question

Let \(B = \{5, 10, 15\}\). Which statements are true: \(\quad 10 \in B, \quad \{5, 15\} \subseteq B, \quad \{10\} \in B, \quad \emptyset \subseteq B\)?

  • \(10 \in B\): true, \(10\) is an element
  • \(\{5, 15\} \subseteq B\): true, both elements of the left set are in \(B\)
  • \(\{10\} \in B\), false: the elements of \(B\) are numbers, not sets (\(\{10\} \subseteq B\) would be true)
  • \(\emptyset \subseteq B\): true, the empty set is a subset of every set

Cardinality

The cardinality \(|A|\) is the number of elements in \(A\):

  • \(|\{\text{road}, \text{rail}, \text{waterway}, \text{ocean}, \text{air}\}| = 5\)
  • \(|\emptyset| = 0\)
  • \(\mathbb{N}\) has infinitely many elements: its cardinality is not a natural number

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.

Your Turn: Count the Elements

Question

What is \(\;|\{1, 2, 2, 3, \{4, 5\}\}|\;\)?

(a) \(3\) (b) \(4\)
(c) \(5\) (d) \(6\)

(b): the elements are \(1\), \(2\), \(3\) and the set \(\{4, 5\}\), four in total.

Common Mistake

The repeated \(2\) counts once (no duplicates in a set), and \(\{4, 5\}\) is one element, not two: it is a set sitting inside a set.

Sets You Already Know

Session 1 was full of sets, we just did not call them that:

  • The number sets \(\mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R}\) are sets
  • Intervals are sets in set-builder notation: \([0, 500] = \{x \in \mathbb{R} : 0 \leq x \leq 500\}\)
  • \(\mathbb{Q} = \{\frac{p}{q} : p, q \in \mathbb{Z}, q \neq 0\}\): set-builder again
  • Today we learn to combine and compare such sets

Set Operations

The Universal Set

Set operations play out inside a universal set \(U\), all objects currently under discussion:

  • Analysing your customer base? \(U\) = all customers of the company
  • Talking about order quantities? \(U = \mathbb{R}\) or \(U = \mathbb{N}_0\)
  • \(U\) is a modelling choice: state it before you start
  • We need \(U\) in a moment to say what “everything not in \(A\)” means

Venn Diagrams

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.

Union

\(A \cup B = \{x : x \in A \text{ or } x \in B\}\): 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.

Intersection

\(A \cap B = \{x : x \in A \text{ and } x \in B\}\): only the elements in both sets.

\(\{1, 2\} \cap \{2, 3\} = \{2\}\): think of it as applying two filters at once.

Set Difference

\(A \setminus B = \{x : x \in A \text{ and } x \notin B\}\): 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.

Complement

\(A^c = \{x \in U : x \notin A\}\): 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: \(A \setminus B = A \cap B^c\).

Four Operations at Once

Let \(U = \{1, 2, \dots, 8\}\), \(A = \{1, 2, 3, 4\}\) and \(B = \{3, 4, 5, 6\}\):

Expression Result Reading
\(A \cup B\) \(\{1, 2, 3, 4, 5, 6\}\) in \(A\) or in \(B\)
\(A \cap B\) \(\{3, 4\}\) in both
\(A \setminus B\) \(\{1, 2\}\) in \(A\), not in \(B\)
\(B \setminus A\) \(\{5, 6\}\) in \(B\), not in \(A\)
\(A^c\) \(\{5, 6, 7, 8\}\) in \(U\), not in \(A\)

Note \(A \setminus B \neq B \setminus A\), and \(A^c\) needs \(U\): without a universal set, “everything not in \(A\)” is undefined.

Your Turn: Compute the Operations

Question

Let \(U = \{1, 2, \dots, 10\}\), \(A = \{2, 4, 6, 8, 10\}\) and \(B = \{1, 2, 3, 4, 5\}\). Shout: \(\;A \cap B, \quad A \setminus B, \quad B^c\), and the number \(|A \cup B|\).

  • \(A \cap B = \{2, 4\}\): the even numbers up to \(5\)
  • \(A \setminus B = \{6, 8, 10\}\): remove \(2\) and \(4\) from \(A\)
  • \(B^c = \{6, 7, 8, 9, 10\}\): everything in \(U\) above \(5\)
  • \(|A \cup B| = |\{1, 2, 3, 4, 5, 6, 8, 10\}| = 8\)

Disjoint Sets

\(A\) 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.

Your Turn: True or False?

Question

For any two sets \(A\) and \(B\): is \(\quad A \cup B \subseteq A \quad\) always true?

  • False: the union is usually bigger than \(A\)
  • \(A\) = premium customers, \(B\) = churned customers: \(A \cup B\) contains churned non-premium customers, who are not in \(A\)
  • The reverse always holds: \(A \subseteq A \cup B\)
  • \(A \cup B \subseteq A\) is only true in the special case \(B \subseteq A\)

How Many in the Union?

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 \((A \cup B)^c\).

Your Turn: Who Bought Nothing?

Question

Of \(200\) customers, \(120\) ordered online, \(90\) ordered in store, and \(40\) did both. How many customers ordered nothing this year? Shout the number.

\[|A \cup B| = 120 + 90 - 40 = 170, \qquad \text{so } 200 - 170 = 30 \text{ ordered nothing.}\]

Common Mistake

\(120 + 90 = 210 > 200\) is not a contradiction: the \(40\) two-channel customers were counted twice. Subtract the intersection before taking the complement.

De Morgan’s Laws for Sets

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.

One Pattern, Two Languages

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.

Your Turn: Late or Incomplete

Question

Let \(A\) = shipments on time, \(B\) = shipments complete. Which set is “shipments that are late or incomplete”?

(a) \(A^c \cup B^c\) (b) \(A^c \cap B^c\)
(c) \((A \cup B)^c\) (d) \(A \cap B\)

(a): late is \(A^c\), incomplete is \(B^c\), and “or” is \(\cup\). By De Morgan this equals \((A \cap B)^c\): everything that is not “on time and complete”.

Common Mistake

  1. and (c) are the same set: shipments that are late and incomplete, a much smaller group. Every “or” of complaints is a \(\cup\).

Products & Power Sets

Ordered Pairs

An ordered pair \((a, b)\) is not a set. Here, order does matter:

  • \((a, b) \neq (b, a)\) unless \(a = b\), but \(\{a, b\} = \{b, a\}\) always
  • The route (Hamburg, Munich) is not the route (Munich, Hamburg)
  • Coordinates in the plane are ordered pairs: \((2, 5) \neq (5, 2)\)

Curly braces \(\{\,\}\) = unordered set, round parentheses \((\,)\) = ordered pair. The bracket type carries the meaning.

The Cartesian Product

\(A \times B = \{(a, b) : a \in A, b \in B\}\): 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

\[|A \times B| = |A| \cdot |B|\]

A Product as a Table

Rows from \(A = \{1, 2\}\), columns from \(B = \{x, y\}\): one ordered pair per cell.

\(x\) \(y\)
\(1\) \((1, x)\) \((1, y)\)
\(2\) \((2, x)\) \((2, y)\)
  • Rows times columns: \(|A| \cdot |B|\) cells, that is where the counting rule comes from
  • Session 1’s cost table \(c_{ij}\) is such a grid: one entry per pair \((i, j)\)
  • Swap the roles and you get \(B \times A\): as many pairs, but \((x, 1) \neq (1, x)\)

Products Build Route Sets

Let \(W = \{1, 2\}\) be warehouses and \(C = \{1, 2, 3\}\) be customers:

  • \(W \times C\) = the set of all possible routes \((i, j)\): here \(2 \cdot 3 = 6\) of them
  • Session 1’s \(x_{ij}\) is defined for every pair \((i, j) \in W \times C\): the index set of the transportation problem is a Cartesian product
  • \(\mathbb{R} \times \mathbb{R} = \mathbb{R}^2\) is the plane: every graph you will draw in Session 3 lives there

Your Turn: How Many Routes?

Question

A carrier runs \(4\) depots and serves \(5\) customers. How many depot-customer routes exist? And if each route can be driven by one of \(2\) vehicle types, how many route-vehicle combinations? Shout both numbers.

  • Routes: \(|D \times C| = 4 \cdot 5 = 20\)
  • With vehicles: \(20 \cdot 2 = 40\), every route combines with every vehicle type
  • The counting rule stacks: three sets give \(|D| \cdot |C| \cdot |V|\) triples \((i, j, v)\)

The Power Set

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 \(1 \notin \mathcal{P}(A)\).

Your Turn: In the Power Set?

Question

Let \(A = \{a, b\}\). Which statement is false?

(a) \(\emptyset \in \mathcal{P}(A)\) (b) \(\{a\} \in \mathcal{P}(A)\)
(c) \(a \in \mathcal{P}(A)\) (d) \(A \in \mathcal{P}(A)\)

(c): \(a\) is an element of \(A\), not a subset of \(A\). Only sets live in \(\mathcal{P}(A)\), and \(\{a\}\) is the set that does.

  1. and (d) are true for every set: \(\emptyset\) and \(A\) itself are always subsets of \(A\).

How Many Subsets?

\[|\mathcal{P}(A)| = 2^{|A|}\]

  • Why: build a subset by deciding for each element: in or out?
  • Two choices per element, \(|A|\) elements: \(2 \cdot 2 \cdot \dots \cdot 2 = 2^{|A|}\)
  • Business reading: subsets are selections: which suppliers to contract, which products to bundle
  • That count grows fast: 10 candidate suppliers already give \(2^{10} = 1024\) possible selections

Your Turn: Counting Options

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.

Sets in Business

Customer Segmentation

Let \(P\) = premium customers, \(C\) = churned customers, \(N\) = newsletter subscribers:

  • \(P \cap C\), premium customers we lost: the win-back campaign list
  • \(C \setminus N\), churned and not reachable by newsletter: needs a phone call
  • \(P^c\), all non-premium customers: the upsell targets
  • Every data filter you will ever build is a set operation: SQL’s AND/OR/NOT are \(\cap\), \(\cup\), \(^c\)

Segments as a Partition

A partition splits \(U\) into disjoint pieces that together cover everything:

  • Regions \(R_1 = \text{North}\), \(R_2 = \text{East}\), \(R_3 = \text{South}\) with \(R_1 \cup R_2 \cup R_3 = U\) and \(R_i \cap R_j = \emptyset\) for \(i \neq j\)
  • Every customer sits in exactly one region: no double counting, nobody missed
  • Then the cardinalities simply add up: \(|U| = |R_1| + |R_2| + |R_3|\), no inclusion-exclusion needed

Overlapping segments (premium and churned) need inclusion-exclusion; a partition is exactly the case where the correction term vanishes.

Reading a Set Expression

How to read \(\;(P \cap C) \setminus N\;\)? Work inside out:

  • \(P \cap C\): customers who are premium and churned
  • \(\dots \setminus N\): now remove everyone on the newsletter
  • Result: lost premium customers we cannot reach by newsletter

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.

Your Turn: Build the Target List

Question

\(P\) = premium, \(C\) = churned, \(N\) = newsletter subscribers. Which expression describes “active premium customers who subscribe to the newsletter”?

(a) \(P \cap C^c \cap N\) (b) \(P \cup C^c \cup N\)
(c) \((P \cap C) \setminus N\) (d) \(P \setminus (C \cap N)\)

(a): premium and not churned and subscribed, three filters at once, so three intersections.

Common Mistake

  1. removes only customers who are churned and subscribed: a churned premium customer without the newsletter stays on the list. Negate the whole condition, not half of it.

Feasible Sets in Optimisation

Every constraint defines a set of allowed values, and constraints combine by intersection:

  • Order quantity \(q\): capacity allows \(q \leq 500\), so \(q \in [0, 500]\)
  • The supplier requires a minimum order of \(20\): \(q \in [20, \infty)\)
  • The feasible set is the intersection: \([0, 500] \cap [20, \infty) = [20, 500]\)
  • Optimisation = picking the best point of the feasible set
  • If the intersection is \(\emptyset\), the problem is infeasible: no plan satisfies all constraints

Your Turn: Still Feasible?

Question

Capacity allows \(q \in [0, 500]\), the supplier requires \(q \in [20, \infty)\). A new customs rule caps single orders at \(15\) units: \(q \in (-\infty, 15]\). What is the feasible set now?

\[[0, 500] \cap [20, \infty) \cap (-\infty, 15] = [20, 500] \cap (-\infty, 15] = \emptyset\]

The problem is infeasible: no order size satisfies all three rules. One constraint has to go, and set notation tells you which pair collides (\(q \geq 20\) against \(q \leq 15\)).

Sets in Your M.Sc. Courses

  • Data Science: events in probability are sets: \(P(A \cup B)\) uses today’s inclusion-exclusion rule
  • Transportation & Distribution: decision variables are indexed by Cartesian products; feasible regions are intersections
  • Analytical Methods: solution sets of equations and inequalities
  • Whenever a slide says “for all \(x \in S\)”, you now read it fluently

Closing

Key Takeaways

  • A set is an unordered collection of distinct elements: \(\in\) for elements, \(\subseteq\) for sets
  • \(\cup\), \(\cap\), \(^c\) mirror or, and, not: De Morgan works in both worlds
  • Venn diagrams turn set expressions into pictures
  • \(|A \cup B| = |A| + |B| - |A \cap B|\): don’t count the intersection twice
  • \(A \times B\) collects ordered pairs; \(\mathcal{P}(A)\) collects subsets, \(|\mathcal{P}(A)| = 2^{|A|}\)
  • Constraints are sets: feasibility is their intersection

That’s it for today.

Any Questions?

Until the Next Session

  • Work through the Tasks: problems with worked solutions
  • Check yourself with the Self-Test quiz
  • Keep the Cheatsheet next to you while practising
  • Note down anything unclear: we start next session with your questions

Draw a Venn diagram whenever a set expression looks confusing: two circles resolve most doubts faster than staring at symbols.

Preview: Session 03: Functions

  • Functions map between sets: every input from one set gets exactly one output in another
  • Domain and range are, you guessed it, sets
  • Linear, quadratic and exponential functions in business
  • Graphs live in \(\mathbb{R} \times \mathbb{R}\), which you met today

See you there, and bring your questions!

Literature & Further Reading

  • These sessions cover the essentials: textbooks offer more depth and practice
  • Start with Sydsæter et al. (2012) or Jacques (2015); full recommendations on the tutorial’s literature page
Jacques, Ian. 2015. Mathematics for Economics and Business. 8th ed. Always Learning. Pearson.
Sydsæter, Knut, Peter J. Hammond, and Arne Strøm. 2012. Essential Mathematics for Economic Analysis. 4th ed. Pearson.