Question

for each pair of statements, choose the one that is true. (a) \\(\circ f \in \\{e, f\\}\\) \\(\circ \\{f\\} \in \\{e, f\\}\\) (b) \\(\circ \\{4\\} \subseteq \\{3, 4\\}\\) \\(\circ 4 \subseteq \\{3, 4\\}\\) (c) \\(\circ j \
ot\in \\{k, l, m\\}\\) \\(\circ j \subseteq \\{j, k, l\\}\\) (d) \\(\circ \\{6, 7, 8\\} \in \\{1, 2, 3, 4, \ldots\\}\\) \\(\circ \\{6, 7, 8\\} \subseteq \\{1, 2, 3, 4, \ldots\\}\\)

Explanation:

Step1: Analyze set membership/subset (a)

• $f \in \{e,f\}$ is false: $f$ is an element, but the notation $\in$ is for elements, but the option uses a set $\{f\}$ in the wrong relation. $\{f\} \subseteq \{e,f\}$ is true because every element of $\{f\}$ is in $\{e,f\}$.

Step2: Analyze set membership/subset (b)

• $4 \subseteq \{3,4\}$ is false: $4$ is an element, not a set, so subset notation does not apply. $\{4\} \subseteq \{3,4\}$ is true because the element 4 of $\{4\}$ is in $\{3,4\}$.

Step3: Analyze set membership/subset (c)

• $j \subseteq \{j,k,l\}$ is false: $j$ is an element, not a set, so subset notation does not apply. $j

otin \{k,l,m\}$ is true because $j$ is not an element of $\{k,l,m\}$.

Step4: Analyze set membership/subset (d)

• $\{6,7,8\} \in \{1,2,3,4,...\}$ is false: $\{6,7,8\}$ is a set, not an element of the natural numbers set. $\{6,7,8\} \subseteq \{1,2,3,4,...\}$ is true because all elements 6,7,8 are in the natural numbers set.

Answer:

(a) $\{f\} \subseteq \{e,f\}$
(b) $\{4\} \subseteq \{3, 4\}$
(c) $j
otin \{k, l, m\}$
(d) $\{6, 7, 8\} \subseteq \{1, 2, 3, 4, ...\}$