Mathematical-Reasoning · Unit 15 · Video 3 · Interactive Practice

Proper Subsets, the Empty Set and Set Equality

IKey Formulas

Statement	Name
$A \subset B$ means $A \subseteq B$ and $A \ne B$	Proper subset
$A \not\subseteq B$ means $(\exists x \in U)[(x \in A) \wedge (x \notin B)]$	Not a subset
For any set $B$ : $\emptyset \subseteq B$ and $B \subseteq B$	Empty set and the set itself
$A = B$ if and only if $A \subseteq B$ and $B \subseteq A$	Test for set equality

Key Insight: $\subseteq$ works like $\le$ and $\subset$ like $<$ . The element symbol relates an object to a set and the subset symbols relate a set to a set: $5 \in B$ , while $\{5\} \subseteq B$ .

IISubset, Proper Subset or Equal?

Checking $A \subseteq B$ and $B \subseteq A$ separately decides whether $A \subset B$ , $B \subset A$ , $A = B$ , or neither.

Start from

IIIA Subset Is Six Conditionals

Over $U = \{1, 2, 3, 4, 5, 6\}$ , $A \subseteq B$ claims $(x \in A) \to (x \in B)$ for every $x$ ; one false row is a witness.

A B

IVFill In Every Symbol That Fits

Each blank takes every symbol of its kind that makes the statement true, and a set equal to another is not a proper subset of it.

$U = \{1, 2, 3, 4, 5, 6\}$ , $A = \{1, 2, 4\}$ , $B = \{1, 2, 3, 5\}$ , $C = \{x \in U \mid x^2 \le 2\}$

With a number on the left the symbols are $\in$ and $\notin$ ; with a set on the left they are $\subset$ , $\subseteq$ , $=$ and $\ne$ . This is the convention the answers follow: read literally, $A \notin B$ and $5 \ne B$ are true as well, since the elements of $B$ are numbers, but those symbols are not of the statement's kind.

Statement

VQuiz Questions

Problem 1 · Proper Subset

Given: $P = \{2, 4\}$ and $Q = \{1, 2, 3, 4\}$ — which statement is true?

✅ Correct! Every element of

P

is in

Q

, and

1 \in Q

with

1 \notin P

, so

P \subseteq Q

and

P \ne Q

❌ Not quite.

1 \in Q

and

1 \notin P

, so that one element shows

Q \not\subseteq P

❌ Not quite. Equal sets have precisely the same elements, but

1 \in Q

and

1 \notin P

❌ Not quite.

P \not\subseteq Q

needs an element of

P

outside

Q

; both

2

and

4

are in

Q

Show solution

$2 \in Q$ and $4 \in Q$ , so every element of $P$ is an element of $Q$ : $P \subseteq Q$ .

$1 \in Q$ and $1 \notin P$ (and also $3 \in Q$ , $3 \notin P$ ), so $P \ne Q$ and $Q \not\subseteq P$ .

Since $P \subseteq Q$ and $P \ne Q$ , the set $P$ is a proper subset of $Q$ : $P \subset Q$ .

Problem 2 · Negating the Subset Relation

Given: $A$ and $B$ are subsets of a universal set $U$ — which statement means $A \not\subseteq B$ ?

✅ Correct! The negation turns

\forall

into

\exists

, and

\neg(P \to Q)

becomes

P \wedge \neg Q

❌ Not quite. This says no element of

A

is in

B

. With

A = \{1, 2\}

and

B = \{2, 3\}

we have

A \not\subseteq B

, yet

2 \in A

and

2 \in B

❌ Not quite. This has the roles of the sets swapped: it is the meaning of

B \not\subseteq A

❌ Not quite. The negation of a conditional is

P \wedge \neg Q

, not another conditional. This statement is true as soon as some

x \in U

is outside

A

, even when

A \subseteq B

Show solution

A \subseteq B \text{ means } (\forall x \in U)[(x \in A) \to (x \in B)].

Negate the whole statement, then move the negation inside the quantifier ( $\neg\forall$ becomes $\exists\neg$ ):

A \not\subseteq B \text{ means } (\exists x \in U)\,\neg[(x \in A) \to (x \in B)].

Since $\neg(P \to Q) \equiv P \wedge \neg Q$ :

A \not\subseteq B \text{ means } (\exists x \in U)[(x \in A) \wedge (x \notin B)].

So one element of $A$ that lies outside $B$ is enough.

Problem 3 · Every Symbol That Fits

Given: $U = \{1, 2, 3, 4, 5, 6, 7, 8\}$ , $D = \{x \in U \mid x \text{ is even}\}$ and $E = \{2, 4\}$ . Each statement has a set on the left, so the candidate symbols are $\subset$ , $\subseteq$ , $=$ and $\ne$ — find every symbol that makes each statement true.

$\{8, 4, 6, 2\}$ ___ $D$

$E$ ___ $\{x \in U \mid x^2 < 20\}$

✅ Correct!

\{8, 4, 6, 2\} = D

, so

\subseteq

and

=

fit; and

E = \{2, 4\}

sits inside

\{1, 2, 3, 4\}

without being equal to it, so

\subset

\subseteq

and

\ne

fit.

❌ Not quite. Order does not matter in a set:

\{8, 4, 6, 2\}

has exactly the elements of

D = \{2, 4, 6, 8\}

, so the two are equal and the proper-subset symbol does not fit.

❌ Not quite.

D = \{2, 4, 6, 8\}

, which has exactly the elements of

\{8, 4, 6, 2\}

; the sets are equal, so neither

\subset

nor

\ne

fits.

❌ Close. The sets are equal, but every set is a subset of itself, so

\subseteq

fits as well.

❌ Close. A proper subset is in particular a subset, so whenever

\subset

fits,

\subseteq

fits too.

❌ Not quite.

\{x \in U \mid x^2 < 20\} = \{1, 2, 3, 4\}

, and

1

is missing from

E

, so the sets are not equal.

❌ Not quite.

\{x \in U \mid x^2 < 20\} = \{1, 2, 3, 4\}

contains both

2

and

4

, so

E

is a subset of it.

Show solution

First statement. $D = \{2, 4, 6, 8\}$ . The set $\{8, 4, 6, 2\}$ has precisely the same elements, since order does not matter, so $\{8, 4, 6, 2\} = D$ . Every set is a subset of itself, so $\{8, 4, 6, 2\} \subseteq D$ . Because the sets are equal, it is not a proper subset. The symbols that fit are $\subseteq$ and $=$ .

Second statement. $1^2 = 1$ , $2^2 = 4$ , $3^2 = 9$ and $4^2 = 16$ are less than $20$ , while $5^2 = 25$ and the larger squares are not, so $\{x \in U \mid x^2 < 20\} = \{1, 2, 3, 4\}$ . Both $2$ and $4$ are in this set, so $E \subseteq \{1, 2, 3, 4\}$ ; and $1$ is in it but not in $E$ , so the sets are not equal. The symbols that fit are $\subset$ , $\subseteq$ and $\ne$ .

Problem 4 · A Subset of the Empty Set

Given: a set $A$ with $A \subseteq \emptyset$ — what can you conclude?

✅ Correct!

A \subseteq \emptyset

is given and

\emptyset \subseteq A

always holds, so the test for set equality gives

A = \emptyset

❌ Not quite.

A \subset \emptyset

would require

A \ne \emptyset

, so

A

would have an element; that element would have to be in

\emptyset

, which has no elements.

❌ Not quite. If

A

had an element

x

, then

x \in A

and

x \notin \emptyset

, which would show

A \not\subseteq \emptyset

. So

A

has no elements.

❌ Not quite.

\emptyset \subseteq A

holds for every set

A

: there is no

x \in \emptyset

with

x \notin A

Show solution

We are given $A \subseteq \emptyset$ .

The empty set is a subset of every set, so $\emptyset \subseteq A$ : the statement $\emptyset \not\subseteq A$ would need an $x \in \emptyset$ with $x \notin A$ , and no such $x$ exists.

Now $A \subseteq \emptyset$ and $\emptyset \subseteq A$ , so by the test for set equality, $A = \emptyset$ .

In particular $A \subset \emptyset$ is impossible, because it would require $A \ne \emptyset$ .

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