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Q3(vii):
Let $A = \{ 1, 2, \{ 3, 4 \}, 5 \}$. Which of the following statements are incorrect and why? (vii) \{1, 2, 5\} $\in$ $A$

Solution :

Given Set Definition

The set $A$ is defined as $A = \{ 1, 2, \{ 3, 4 \}, 5 \}$.

To determine the correctness of the statement $\{1, 2, 5\} \in A$, we must analyze the membership criteria of set $A$ based on the definition of the element-of relation ($\in$).

Step 1: Identifying Elements of Set $A$

The elements of set $A$ are those objects explicitly separated by commas within the set brackets. By inspection:

  • $1$ is an element of $A$ ($1 \in A$)
  • $2$ is an element of $A$ ($2 \in A$)
  • $\{3, 4\}$ is an element of $A$ ($\{3, 4\} \in A$)
  • $5$ is an element of $A$ ($5 \in A$)

[Per the definition of Set Membership, an object $x$ is an element of $A$ if and only if $x$ is listed as a primary member of the set.]

Step 2: Evaluating the Statement $\{1, 2, 5\} \in A$

The statement posits that the set $\{1, 2, 5\}$ is an element of $A$. For this to be true, the object $\{1, 2, 5\}$ must appear as an explicit entry within the curly braces of $A$. Comparing the set $\{1, 2, 5\}$ against the list of elements identified in Step 1, we observe that it is not present.

Step 3: Distinguishing Between Membership ($\in$) and Subset ($\subset$)

It is crucial to note the difference between being an element of a set and being a subset of a set:

Relation Definition Application to $\{1, 2, 5\}$
Element ($\in$) The object must be a direct member of the set. $\{1, 2, 5\} \notin A$ because it is not an element.
Subset ($\subset$) Every element of the set must be present in the parent set. $\{1, 2, 5\} \subset A$ is true because $1 \in A, 2 \in A,$ and $5 \in A$.

[By the Principle of Set Inclusion, $B \subset A$ if $\forall x (x \in B \implies x \in A)$. Since $1, 2, 5 \in A$, the subset relation holds, not the element relation.]

1 2 {3,4} 5 Set A

Final Solution: The statement $\{1, 2, 5\} \in A$ is incorrect. Although $1, 2,$ and $5$ are individual elements of $A$, the set $\{1, 2, 5\}$ itself is not listed as an element within $A$. It is, however, a proper subset of $A$ ($\{1, 2, 5\} \subset A$).


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