in Set Theory & Algebra edited by
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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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4 Comments

An element of a set is never a subset of the set. For that the element must be inside a set

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6
Does  this phi in 2nd statement is  a set represents by { }  ?
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Yes @Golam Murtuza

But why $\varnothing $ or $\left\{ \right\}$ is a subset of  any set?

Set $A$ is subset of set $B$ iff  $$ \forall x \left(  x \in A \to x \in B \right)$$

Let $A=\varnothing $since no element can belong to empty set, $x \in A$ will be always $false$

Since $false \to anything$ is $true$  (Vacuos truth ) we can say $\varnothing $ is subset of any set. 

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6 Answers

117 votes
117 votes
Best answer
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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4 Comments

@Rutuja

Yes, there is a difference.

ϕ == {} (an empty set)

=> An empty set is a subset of any set => it is a subset of the powerset too

{ϕ} == a set containing ϕ

=> {ϕ} is a subset of the powerset because ϕ is present in the powerset so we can form a subset with it.

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@

thankss!

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@Arjun sir i think there is a edit needed in last. To make option IV true 5,{6} should be the elements of 2^A so that {5,{6}} ⊆ 2^A which is never possible because 5 is not even a set and 2^A contains only sets.

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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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