   Question Answers

# Write down all the subsets of the following sets $\{ 1,2,3,4,5,.....\infty \}$1) $\{ a\}$2) $\{ a,b\}$3) $\{ 1,2,3\}$4) $\emptyset$

Hint: If $A$ and $B$ are two sets, we define the subset “if $A$ is a subset of $B$ then, all members of $A$ is a member of $B$”.Subset is denoted by $A \subseteq B$ and $\emptyset$ is the empty set.
Here we going to write subsets, for that, we use some rules,
The first rule is, “Empty set is a subset of every set”.
The second rule is, “The set itself is a subset of the set”.

1) $\{ a\}$
We can take $A = \{ a\}$
As we say the first rule, “empty set is a subset of every set”
That is
$\emptyset \subseteq A$
As we say the second rule, “the set itself is a subset for set”
$\{ a\} \subseteq A$
Hence $\emptyset ,\{ a\}$ are the only subsets of $A = \{ a\}$

2) $\{ a,b\}$
We can take $A = \{ a,b\}$
As we say the first rule
$\emptyset \subseteq A$
The second rule,
$\{ a,b\} \subseteq A$
Also we can write,
$\{ a\} \subseteq A$
$\{ b\} \subseteq A$
Hence $\emptyset ,\{ a,b\} ,\{ a\} ,\{ b\}$ are the subsets of set $A = \{ a,b\}$

3) $\{ 1,2,3\}$
We can take $A = \{ 1,2,3\}$
By the first rule,
$\emptyset \subseteq A$
In the second rule,
$\{ 1,2,3\} \subseteq A$
Then,
$\{ 1\} \subseteq A$
$\{ 2\} \subseteq A$
$\{ 3\} \subseteq A$
Hence $\emptyset ,\{ 1,2,3\} ,\{ 1\} ,\{ 2\} ,\{ 3\}$ are the subsets of set $A = \{ 1,2,3\}$

4) $\emptyset$
It is different from other sets, $\emptyset$ is the empty set.
Empty set has no elements inside the set.
Therefore it has no subsets.
Hence the set $\emptyset$ has no subsets.

Note:The rule “Empty set is subset for every set” is called trivial subset.And the rule “the set itself is a subset of a set” is called improper subset.Here we can observe that the section (iv) has $\emptyset$ is empty set it has no elements, But we can apply the first rule $\emptyset$ is the subset for the set $\emptyset$ ($\emptyset \subseteq \emptyset$) and also we apply the second rule $\emptyset$ is the subset for the set $\emptyset$ ($\emptyset \subseteq \emptyset$). Hence both rules are equal also it has no elements.