Set Theory: P(P{1}) and its cardinality

  • #1
The notation has me a bit confused...

Heres my logic

for the P({1}) on the inside

{EmptySet, {{1}}}
reason being, you always include the empty set, {1} is a part of the set. The cardinality is two

You have the set: {EmptySet, {{1}}}, and now you have to consider the outer "P"
the possibilities are:{EmptySet, {{1}}},EmptySet, {{1}}
I feel like there should be one more as its cardinality is three, when it should be four, and I dont think order matters
 

Answers and Replies

  • #2
795
7
The notation has me a bit confused...

Heres my logic

for the P({1}) on the inside

{EmptySet, {{1}}}
reason being, you always include the empty set, {1} is a part of the set. The cardinality is two
You're fine except that you're mistaken about thinking the empty set is an element of every set. What's true is that the empty set is a subset of every set. So the empty set is an element of the power set of every set.

∅ is the empty set.

{∅} is the set containing the empty set. It's the set we call '1' in the von Neumann ordinals. In other words,

{∅} = 1.

Then {∅, 1} = {∅, {∅}} = 2 and so forth. Note that the number of elements of each set is equal to the name of the set. 1 has 1 element, 2 has 2 elements.

Remember the distinction between an element of a set and a subset of a set. The empty set has no elements, but it does have one subset, namely the empty set. The empty set is a subset of every set; but the empty set may or may not be an element of some other given set.
 

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