Discrete Math for Data Science

DSCI 220, 2025 W1

October 27, 2025

Announcements

Sets Continued

Goals

• Gallery through proofs

Warm Up

Suppose \(A = \{1, \{2\}\}\)

1. Is \(1\in A\)?
2. Is \(\{1\} \subseteq A\)?
3. Is \(1 \subseteq A\)?
4. Is \(\{1\} \in A\)?
5. Is \(2\in A\)?

6. Is \(\{2\} \subseteq A\)?
7. Is \(2 \subseteq A\)?
8. Is \(\{2\} \in A\)?
9. Is \(\emptyset \in A\)?
10. Is \(\emptyset \subseteq A\)?

Set Proofs

Setup

We’ll use these café sets:

\(I\) = “iced orders” ( df['is_iced'])

\(N\) = “non-dairy milk” (df['milk']=='Oat' | df['milk'] == 'Almond')

\(O\) = “oat-milk orders” (df['milk'] == 'Oat')

\(A\) = “almond-milk orders” (df['milk'] == 'Almond')

\(H\) = “high-caffeine” (df['caffeine_mg'] >= 150)

\(L\) = “low-calorie” (df['calories'] <= 150)

DeMorgan for Sets

By observation, you discovered that \((I\cup N)^c = I^c\cap N^c\). But is this always true?

 

Claim: For any sets \(A\) and \(B\), \((A\cup B)^c = A^c\cap B^c\).

 

Observations, before we begin:

• universally quantified statement requires arbitrary instantiation of variables.
• set equality \(A=B\) actually requires 2 proofs: \(A\subseteq B\) and \(B\subseteq A\).
• to prove \(A\subseteq B\) we rely on the definition: For any \(x\), \(x\in A\rightarrow x\in B\).
• following the form of the definition, we consider an arbitrary \(x\), assume \(x\in A\) and show \(x\in B\).

Union Widens

Story: Broadening a filter to “non-dairy or iced” can’t lose non-dairy orders.

Theorem: \(N \subseteq N\cup I\).

Proof:

 

 

Apply:

• \(|N|\le |N\cup I|\)
• every non-dairy row still passes the broader filter.

Union is Monotone

Story: Every oat drink is non-dairy. Then “oat or high-caffeine” is contained in “non-dairy or high-caffeine”.

Theorem: If \(O\subseteq N\), then \(O\cup H \subseteq N\cup H\).

Proof: Let \(x\in O\cup H\). Then \(x\in H\) or \(x\in O\).

• If \(x\in H\), then \(x\in N\cup H\).
  
• If \(x\in O\), then \(x\in N\) (since \(O\subseteq N\)), so \(x\in N\cup H\).

Apply: Replacing a condition by a superset inside an OR cannot shrink results.

Complement Flips the Order

Story: Since oat \(\subseteq\) non-dairy, anyone not non-dairy is certainly not oat.

Theorem: If \(O\subseteq N\), then \(N^{c}\subseteq O^{c}\).

Proof:

 

 

Apply: Negating a broader class yields a subset in the reverse direction.

Take \(x\in N^c\) so \(x\notin N\). If \(x\in O\) then (because \(O\subseteq N\)) we’d have \(x\in N\)—contradiction. Hence \(x\notin O\), i.e., \(x\in O^c\). \(\square\)

Partition Identity

Story: Split iced orders by whether they’re non-dairy (\(N\)) or not (\(N^c\)). We should neither lose nor double-count any iced orders.

Theorem: For any sets \(I\) and \(N\), \[ (I\cap N)\ \cup\ (I\cap N^{c}) \;=\; I \quad\text{and}\quad (I\cap N)\ \cap\ (I\cap N^{c}) \;=\; \varnothing. \]

Apply:

• Because the union is “disjoint”, \(|I| \;=\; |I\cap N| \;+\; |I\cap N^c|\)
• This is the basis for the “Law of Total Probability,” which you will see someday!

Partition Identity Proof

Theorem: For any sets \(I\) and \(N\), \[ (I\cap N)\ \cup\ (I\cap N^{c}) \;=\; I \quad\text{and}\quad (I\cap N)\ \cap\ (I\cap N^{c}) \;=\; \varnothing. \]

Proof: