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First of all, is A ={φ} a set? If yes, then what are its subsets?

Note: • φ represents an empty set.

• A must have two distinct subsets, (∵ n(A)=1, thus 2¹= 2)

My approach: According to me, if it has two subsets then they should be φ and φ. If it's so then they aren't distinct at all.

Gaurav Kumar
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  • The usual symbol foe empty set is $\emptyset$ (\emptyset). Having said that, yes : $A = { \emptyset }$ is a set with exactly one element: the empty set. – Mauro ALLEGRANZA Apr 17 '20 at 08:48
  • And yes; every finite set with $n$ elements has $2^n$ subsets. Thus, having $A$ one element, the number of its subsets is $2$. Every set $A$ has $\emptyset$ and $A$ itself between its subsets; thus... – Mauro ALLEGRANZA Apr 17 '20 at 08:49
  • In conclusion, $A = { \emptyset }$ has two subsets: $\emptyset$ and ${ \emptyset }$. The two are different: the first one is empty while the second one has one element. – Mauro ALLEGRANZA Apr 17 '20 at 08:57
  • Yes. Thank you so much. – Gaurav Kumar Apr 17 '20 at 09:05

2 Answers2

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Notice that $\{\emptyset\} \ne \emptyset.$

$$\mathcal{P} ( \{\emptyset\} )=\{\emptyset, \{\emptyset\} \}.$$

$$\mathcal{P} ( \emptyset )=\{\emptyset\}.$$

Riccardo
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  • In virtue of the definition of " subset"

in order a set $X$ to be a subset of $\{\emptyset\}$ it is necessary and sufficient that all the elements of set $X$ be also elements of $\{\emptyset\}$.

  • Your question therefore amounts to :

which set(s) can I substitute for $X$ in the sentence " all the elements of set $X$ are also elements of $\{\emptyset\}$ " in order to make a true sentence?

  • Setting $X = \emptyset$ results in a true sentence.

Note : since $\emptyset$ has no element, one can truly say that all its elements belong to any arbitrary set S ; otherwise, it would mean that some element of $\emptyset$ does not belong to S, which is absurd.

  • Setting $X = \{\emptyset\}$ also results in a true sentence.

For, certainly, $\{\emptyset\}$ has no element that does not belong to $\{\emptyset\}$ itself.

  • No other substitution would result in a true sentence.

In fact I already have 2 subsets, and , as you pointed out, since the cardinal of your original set is $1$, the cardnal of its power set ( that is, the number of its subsets) is $2^1 =2$.

  • So $P(\{\emptyset\}) = \{\emptyset, \{\emptyset\}\}$.