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64 votes
64 votes

For a set $A$, the power set of $A$ is denoted by $2^{A}$. If $A = \left\{5,\left\{6\right\}, \left\{7\right\}\right\}$, which of the following options are TRUE?

  1. $\varnothing \in 2^{A}$
  2. $\varnothing  \subseteq 2^{A}$
  3. $\left\{5,\left\{6\right\}\right\} \in 2^{A}$
  4. $\left\{5,\left\{6\right\}\right\} \subseteq 2^{A}$
  1. I and III only
  2. II and III only
  3. I, II and III only
  4. I, II and IV only
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6 Answers

Best answer
117 votes
117 votes
Power set of $A$ consists of all subsets of $A$ and from the definition of a subset, $\emptyset$ is a subset of any set. So, $\text{I}$ and $\text{II}$ are TRUE.

$5$ and $\{6\}$ are elements of $A$ and hence $\{5, \{6\} \}$ is a subset of $A$ and hence an element of $2^{A}$. An element of a set is never a subset of the set. For that the element must be inside a set- i.e., a singleton set containing the element is a subset of the set, but the element itself is not. Here, option $\text{IV}$ is false. To make $\text{IV}$ true we have to do as follows:

$\{5, \{6\} \}$ is an element of $2^{A}$. So, $\{ \{5, \{6\} \}\}\subseteq  2^{A}.$

So, option C.
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28 votes
28 votes

A={5,{6},{7}

2A  = p = {  ϕ, {5}, {{6}}, {{7}}, {5, {6}}, {5, {7}}, {{6}, {7}}, {5, {6}, {7}}  }

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22 votes
22 votes

We use "subset" symbol to compare 2 sets ...Ex: Set A is a subset of set B iff every elements of set A is in set B.

We use "belongs to" symbol to compare a set and an element. Ex: Whether an element is present inside a set or not.

{5,{6}} is not a subset of 2A .. Because 5 is not present in 2A. {5} is actually present in 2A

{5} (a set containing an element 5) is different from 5 (an element)...

5 votes
5 votes

option C

phi is subset of every set.

phi is the first element of the power set

{5,{6}} ⊆ A

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