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Hi I'm working on some basic set/subset comparison statements in the form of True/False. There are 6 statements that I have to determine whether they're true or false. I have the first and last down, but I do not understand why I am wrong about the other four.

If A = {One, {One, Two}, Three, {Three}, Four, {Two}, One}

Determine if the following are true or false.

  1. {Four, {Two}} ∈ A

I said false

  1. {One} ⊆ P(A)

I said True, because in the power set of A, {One} is one of the subsets, but this was wrong apparently.

  1. {One, Two} ⊆ A

I said true because all of the elements in the set on the left are in A, but this was wrong apparently.

  1. {Three} ⊆ A

I said true because {Three} is a subset of A, but this was wrong apparently.

  1. {Three} ∈ A

I said true because {Three} is an element in A, but this was wrong apparently.

  1. {Four, Four, Three} ∈ P(A)

I said false, because two different instances of Four don't exist in A, so it wouldn't be a subset and this appears to be correct.

What am I not understanding conceptually, and why am I wrong for 2-5?

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  • $\begingroup$ {One} is a subset of A (and an element of P(A)), which is not the same as being a subset of P(A). {One, Two} is an element of A, not a subset. It seems to me you are right about 4 and 5. In (usual) set-theory there are no different instances, that is {Four, Four, Three} means the same as {Four, Three} $\endgroup$ – Mirko Oct 25 '19 at 3:18
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A = {One, {One, Two}, Three, {Three}, Four, {Two}, One}

  1. {One} ⊆ P(A)

I said True, because in the power set of A, {One} is one of the subsets, but this was wrong apparently.

I consider your answer correct and the other incorrect.

  1. {One, Two} ⊆ A

I said true because all of the elements in the set on the left are in A, but this was wrong apparently.

2 is not an element of A. 2 is an element of two of the elements of A.

  1. {Three} ⊆ A

I said true because {Three} is a subset of A, but this was wrong apparently.

You are correct.

  1. {Three} ∈ A

I said true because {Three} is an element in A, but this was wrong apparently.

You are correct. Both 3 and {3} are distinct elements of A.

  1. {Four, Four, Three} ∈ P(A)

I said false, because two different instances of Four don't exist in A, so it wouldn't be a subset and this appears to be correct.

Sets do not have multiple entries. Multiset do.
Thus {4, 4, 3} = {4, 3} which is not an element of A but a subset.

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  • $\begingroup$ I understand now, thank you! One thing That has been brought up is that, the reason 2. is false is because {One} ∈ P(A), and {{One}} ⊆ P(A), but {One} is not a subset of the power set, since a subset of the powerset must be a set that has sets as the elements. $\endgroup$ – Caed Oct 25 '19 at 3:51

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