GATE CSE 2015 | Set 1 | Question: 1‏6

Question

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. $\left\{5,\left\{6\right\}\right\} \subseteq 2^{A}$

• A. I and III only
• B. II and III only
• C. I, II and III only
• D. I, II and IV only

Comments

why {5,{6}}⊆2^A  is incoorect

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$\{5, \{6\}\} \in 2^A$
So, $\{\{5, \{6\}\}\} \subseteq 2^A$
$\{5, \{6\}\} \not\subseteq 2^A$

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IV is false, {5, {6}} is not a subset, but {{5, {6}}} is.

What does this mean?

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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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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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Ans. (C)

4 th one is wrong, because to be a subset it needs to be enclosed within one more braces.

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Just apply (∀x)(x ∈ ∅⇒ x ∈ S)

• ∅ ∈ P(S) because it appears as an element inside the power set.
• ∅ ⊆ P(S) because the empty set is a subset of every set.

Answer 1

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.

Comments

@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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Ohh {5,{6}} belongs to P(A)

and {{5,{6}}} is subset of P(A)

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@N thanks for pdf

Answer 2

A={5,{6},{7}} 

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

Answer 3

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 2^A .. Because 5 is not present in 2^A. {5} is actually present in 2^A. 

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

Answer 4

option C

phi is subset of every set.

phi is the first element of the power set

{5,{6}} ⊆ A

Answer 5

A = { 5 , {6} , {7} } 

P(A) = { ∅ , {5} , {{6}} , {{7}} , {5,{6}} , {5,{7}} , {{6},{7}} }
 

I. ∅ ∈ $2^A$ is True , since P(A) contains ∅ .

II. ∅ ⊆  $2^A$ is True because empty set is a subset of every set — by definition of subset.

now , we cannot say that 5 is a subset of A because 5 is an element in A so 5 ∈ A .

5 ⊆  A  ❌  

{5} ⊆  A ✔️

5 ∈  A  ✔️

similarly {5, {6}} is an element of $2^A$ . So, {{5, {6}}} ⊆ $2^A$.

therefore III is True but IV is False.

Finally Option (C) is correct.

Answer 6

II. is trivially true. $\phi$ is a subset of any set.

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I. $ϕ∈2^A$. Means is $\phi$ an element of $2^A\ ?$ Yes, it is. $^{[1]}$

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III. Enumerate the elements of $2^A$. $\{5,\{6\}\}$ is an element in it. So, true.

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IV. Is $\{5,\{6\}\}$ a subset of $2^A\ ?$ This implicitly assumes that $\{5,\{6\}\}$ is a set. So, we can rephrase it as: are the elements $5$ and $\{6\}$ present in $2^A \ ?$

The answer is no.

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Option C

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$^{[1]}$ Power set is the set of all subsets.

Hence, the elements of a powerset are sets in themselves.

$\phi$ is the empty set (it literally is the definition of it), so $\phi$ will always belong $(∈)$ to any power set.