Question 19: $\{V, C, R\}$ $\qquad$ $\{\mathrm{V}, \mathrm{C}, \mathrm{R}, \mathrm{S}\}$
Understanding subset relations
For sets A and B:
- $A \subseteq B$ means every element of A is also in B (A is a subset of B)
- $A \subset B$ means A is a proper subset of B (A is a subset of B and A ≠ B)
For the sets $\{V, C, R\}$ and $\{V, C, R, S\}$, I need to check if all elements of the first set are in the second set.
Analyzing the sets
The first set has elements V, C, and R.
The second set has elements V, C, R, and S.
All elements of the first set (V, C, R) are contained in the second set.
Additionally, the second set has an extra element S that is not in the first set.
Determining the relationship
Since all elements of the first set are in the second set, $\{V, C, R\} \subseteq \{V, C, R, S\}$ is true.
Since the sets are not equal (the second set has an extra element S), $\{V, C, R\} \subset \{V, C, R, S\}$ is also true.
Therefore, both $\subseteq$ and $\subset$ can be placed in the blank to form a true statement.
\(\boxed{\text{both}}\)
Question 20: $\{F, I, N\}$ $\qquad$ \{F, I, N, K \}
Analyzing the sets
The first set has elements F, I, and N.
The second set has elements F, I, N, and K.
All elements of the first set (F, I, N) are contained in the second set.
Additionally, the second set has an extra element K that is not in the first set.
Determining the relationship
Since all elements of the first set are in the second set, $\{F, I, N\} \subseteq \{F, I, N, K\}$ is true.
Since the sets are not equal (the second set has an extra element K), $\{F, I, N\} \subset \{F, I, N, K\}$ is also true.
Therefore, both $\subseteq$ and $\subset$ can be placed in the blank to form a true statement.
\(\boxed{\text{both}}\)
Question 21: $\{0,2,4,6,8\}$ $\qquad$ $\{8,0,6,2,4\}$
Analyzing the sets
The first set has elements 0, 2, 4, 6, and 8.
The second set has elements 8, 0, 6, 2, and 4.
These sets contain exactly the same elements, just listed in a different order. In set theory, the order of elements doesn't matter.
Determining the relationship
Since both sets contain exactly the same elements, they are equal: $\{0,2,4,6,8\} = \{8,0,6,2,4\}$
For equal sets:
- $A \subseteq B$ is true (every element of A is in B)
- $A \subset B$ is false (because A = B, not a proper subset)
Therefore, only $\subseteq$ can be placed in the blank to form a true statement.
\(\boxed{\subseteq}\)
Question 22: $\{9,1,7,3,4\}$ $\qquad$ $\{1,3,4,7,9\}$
Analyzing the sets
The first set has elements 9, 1, 7, 3, and 4.
The second set has elements 1, 3, 4, 7, and 9.
These sets contain exactly the same elements, just listed in a different order. In set theory, the order of elements doesn't matter.
Determining the relationship
Since both sets contain exactly the same elements, they are equal: $\{9,1,7,3,4\} = \{1,3,4,7,9\}$
For equal sets:
- $A \subseteq B$ is true (every element of A is in B)
- $A \subset B$ is false (because A = B, not a proper subset)
Therefore, only $\subseteq$ can be placed in the blank to form a true statement.
\(\boxed{\subseteq}\)
Question 23: $\{x \mid x$ is a man $\}$ $\qquad$ $\{x \mid x$ is a woman $\}$
Analyzing the sets
The first set contains all men.
The second set contains all women.
These are disjoint sets - no man is a woman, and no woman is a man.
Determining the relationship
Since no element of the first set is in the second set, the first set is not a subset of the second set.
Therefore, neither $\subseteq$ nor $\subset$ can be placed in the blank to form a true statement.
\(\boxed{\text{neither}}\)
Question 24: $\{x \mid x$ is a woman $\}$ $\qquad$ $\{x \mid x$ is a man $\}$
Analyzing the sets
The first set contains all women.
The second set contains all men.
These are disjoint sets - no woman is a man, and no man is a woman.
Determining the relationship
Since no element of the first set is in the second set, the first set is not a subset of the second set.
Therefore, neither $\subseteq$ nor $\subset$ can be placed in the blank to form a true statement.
\(\boxed{\text{neither}}\)
Question 25: $\{x \mid x$ is a man $\}$ $\qquad$ $\{x \mid x$ is a person $\}$
Analyzing the sets
The first set contains all men.
The second set contains all persons.
Every man is a person, so every element of the first set is in the second set.
Additionally, there are persons who are not men (e.g., women), so the sets are not equal.
Determining the relationship
Since every element of the first set is in the second set, $\{x \mid x \text{ is a man}\} \subseteq \{x \mid x \text{ is a person}\}$ is true.
Since the sets are not equal (there are persons who are not men), $\{x \mid x \text{ is a man}\} \subset \{x \mid x \text{ is a person}\}$ is also true.
Therefore, both $\subseteq$ and $\subset$ can be placed in the blank to form a true statement.
\(\boxed{\text{both}}\)
Question 26: $\{x \mid x$ is a woman $\}$ $\qquad$ $\{x \mid x$ is a person $\}$
Analyzing the sets
The first set contains all women.
The second set contains all persons.
Every woman is a person, so every element of the first set is in the second set.
Additionally, there are persons who are not women (e.g., men), so the sets are not equal.
Determining the relationship
Since every element of the first set is in the second set, $\{x \mid x \text{ is a woman}\} \subseteq \{x \mid x \text{ is a person}\}$ is true.
Since the sets are not equal (there are persons who are not women), $\{x \mid x \text{ is a woman}\} \subset \{x \mid x \text{ is a person}\}$ is also true.
Therefore, both $\subseteq$ and $\subset$ can be placed in the blank to form a true statement.
\(\boxed{\text{both}}\)
Question 27: $\{x \mid x$ is a man or a woman $\}$ $\qquad$ $\{x \mid x$ is a person $\}$
Analyzing the sets
The first set contains all men and all women.
The second set contains all persons.
In general, all persons are either men or women (in the context of this problem), so these sets contain the same elements.
Determining the relationship
Since both sets contain exactly the same elements, they are equal: $\{x \mid x \text{ is a man or a woman}\} = \{x \mid x \text{ is a person}\}$
For equal sets:
- $A \subseteq B$ is true (every element of A is in B)
- $A \subset B$ is false (because A = B, not a proper subset)
Therefore, only $\subseteq$ can be placed in the blank to form a true statement.
\(\boxed{\subseteq}\)
Question 28: $\{x \mid x$ is a woman or a man $\}$ $\qquad$ $\{x \mid x$ is a person $\}$
Analyzing the sets
The first set contains all women and all men.
The second set contains all persons.
In general, all persons are either women or men (in the context of this problem), so these sets contain the same elements.
Determining the relationship
Since both sets contain exactly the same elements, they are equal: $\{x \mid x \text{ is a woman or a man}\} = \{x \mid x \text{ is a person}\}$
For equal sets:
- $A \subseteq B$ is true (every element of A is in B)
- $A \subset B$ is false (because A = B, not a proper subset)
Therefore, only $\subseteq$ can be placed in the blank to form a true statement.
\(\boxed{\subseteq}\)
Question 29: $A=\{x \mid x \in \mathbf{N}$ and $5<x<12\}$, $B=$ the set of natural numbers between 5 and 12, A $\qquad$ B
Analyzing the sets
Set A = {x | x ∈ ℕ and 5 < x < 12} = {6, 7, 8, 9, 10, 11}
Set B = the set of natural numbers between 5 and 12 = {6, 7, 8, 9, 10, 11}
Both sets contain exactly the same elements.
Determining the relationship
Since both sets contain exactly the same elements, they are equal: A = B
For equal sets:
- $A \subseteq B$ is true (every element of A is in B)
- $A \subset B$ is false (because A = B, not a proper subset)
Therefore, only $\subseteq$ can be placed in the blank to form a true statement.
\(\boxed{\subseteq}\)
Question 30: $A=\{x \mid x \in \mathbf{N}$ and $3<x<10\}$, $B=$ the set of natural numbers between 3 and 10, A $\qquad$ B
Analyzing the sets
Set A = {x | x ∈ ℕ and 3 < x < 10} = {4, 5, 6, 7, 8, 9}
Set B = the set of natural numbers between 3 and 10 = {4, 5, 6, 7, 8, 9}
Both sets contain exactly the same elements.
Determining the relationship
Since both sets contain exactly the same elements, they are equal: A = B
For equal sets:
- $A \subseteq B$ is true (every element of A is in B)
- $A \subset B$ is false (because A = B, not a proper subset)
Therefore, only $\subseteq$ can be placed in the blank to form a true statement.
\(\boxed{\subseteq}\)
Question 31: $A=\{x \mid x \in \mathbf{N}$ and $5<x<12\}$, $B=$ the set of natural numbers between 3 and 17, A $\qquad$ B
Analyzing the sets
Set A = {x | x ∈ ℕ and 5 < x < 12} = {6, 7, 8, 9, 10, 11}
Set B = the set of natural numbers between 3 and 17 = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}
All elements of set A are contained in set B, but set B has additional elements that are not in set A.
Determining the relationship
Since every element of A is in B, $A \subseteq B$ is true.
Since the sets are not equal (B has elements that A doesn't have), $A \subset B$ is also true.
Therefore, both $\subseteq$ and $\subset$ can be placed in the blank to form a true statement.
\(\boxed{\text{both}}\)
Question 32: $A=\{x \mid x \in \mathbf{N}$ and $3<x<10\}$, $B=$ the set of natural numbers between 2 and 16, A $\qquad$ B
Analyzing the sets
Set A = {x | x ∈ ℕ and 3 < x < 10} = {4, 5, 6, 7, 8, 9}
Set B = the set of natural numbers between 2 and 16 = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
All elements of set A are contained in set B, but set B has additional elements that are not in set A.
Determining the relationship
Since every element of A is in B, $A \subseteq B$ is true.
Since the sets are not equal (B has elements that A doesn't have), $A \subset B$ is also true.
Therefore, both $\subseteq$ and $\subset$ can be placed in the blank to form a true statement.
\(\boxed{\text{both}}\)
Question 33: $A=\{x \mid x \in \mathbf{N}$ and $5<x<12\}$, $B=\{x \mid x \in \mathbf{N}$ and $2 \leq x \leq 11\}$, A $\qquad$ B
Analyzing the sets
Set A = {x | x ∈ ℕ and 5 < x < 12} = {6, 7, 8, 9, 10, 11}
Set B = {x | x ∈ ℕ and 2 ≤ x ≤ 11} = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
All elements of set A are contained in set B, but set B has additional elements (2, 3, 4, 5) that are not in set A.
Determining the relationship
Since every element of A is in B, $A \subseteq B$ is true.
Since the sets are not equal (B has elements that A doesn't have), $A \subset B$ is also true.
Therefore, both $\subseteq$ and $\subset$ can be placed in the blank to form a true statement.
\(\boxed{\text{both}}\)
Question 34: $A=\{x \mid x \in \mathbf{N}$ and $3<x<10\}$, $B=\{x \mid x \in \mathbf{N}$ and $2 \leq x \leq 8\}$, A $\qquad$ B
Analyzing the sets
Answered
