# Elements of a set.

## Homework Statement

Write out the following sets by listing their elements between the curly braces.

a. $2^\emptyset$
b. {x:x $\subseteq$ {1,2,3,4,5} and |x|$\leq$ 1

## The Attempt at a Solution

a. I'm not sure if the only element is one, or if this is an undefined operation.
b. I think the elements of this set are ${\emptyset}$ and {1}

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I would say
a) {1}
b) {1}

for part b. I think the empty set should be included because it says that x is a subset of that set and the empty set is a subset of any set.

a. I'm not sure if the only element is one, or if this is an undefined operation.
Given any set S, we can always form its power set. This is just the set of all subsets of S. What are the subsets of the empty set?

b. I think the elements of this set are ${\emptyset}$ and {1}
I think you're misunderstanding the question. |S|≤ 1 means that the cardinality of S is less than or equal to 1, i.e., S has one element or fewer. What subsets can you form with one element or fewer?

ok so the only subset of the empty set is the empty set so it has 1 element.
and then my answer for part b will just have 1 element.

ok so the only subset of the empty set is the empty set so it has 1 element.
Right. So $\mathcal{P}(\emptyset) =$ ?

and then my answer for part b will just have 1 element.
I think you're getting a little mixed up. Remember that elements of the power set are in fact sets themselves!

Let's call $S = \{X : X \subseteq \{1, 2, 3, 4, 5\}$ and $|X| \leq 1\}$. I claim that the set {2} is in S. Do you see why?

Right. So $\mathcal{P}(\emptyset) =$ ?

I think you're getting a little mixed up. Remember that elements of the power set are in fact sets themselves!

Let's call $S = \{X : X \subseteq \{1, 2, 3, 4, 5\}$ and $|X| \leq 1\}$. I claim that the set {2} is in S. Do you see why?

So the power set of the empty set is the empty set.
and {2} is in S because it has just 1 element.

So the power set of the empty set is the empty set.
Not quite. The only element of $\mathcal{P}(\emptyset)$ is $\emptyset$. So $\mathcal{P}(\emptyset)$ can't be empty: I just named one of its elements. Things get tricky when we have sets whose elements are sets whose elements are sets...

So, $\mathcal{P}(\emptyset)$=?
and {2} is in S because it has just 1 element.
Right, so can you write out S explicitly, listing all its members?

so the power set of the empty set has 1 element.

Dick