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Set Theory (Posted on 2014-04-19) Difficulty: 3 of 5

The following is a list of 15 thought provoking statements
about sets where a true or false answer is needed.
A denotes any set, F the set {1,2}, φ the empty set,
and ℘(S) the set of all subsets of set S (i.e, the power
set of S).

a)   φ∩{φ} = {φ}
b)   φ∪{φ} = {φ,{φ}}
c)   φ ∈ ℘({φ,{φ}})
d)   {φ} ⊆ A
e)   φ ⊆ A
f)   φ ⊆ ℘(A)
g)   {{φ}} ⊆ ℘(φ)
h)   {φ}∩φ = φ
i)   ℘(φ) = {φ,{φ}}
j)   φ ∈ A
k)   φ ∈ ℘(A)
l)   {φ} ∈ ℘(A)
m)   {φ}∪φ = {φ}
n)   φ ⊆ ℘(F)-φ
o)   {φ} ⊆ {{φ,{φ},{{φ}}}}

From "Bridge to Abstract Mathematics",1987
          by Ronald P. Morash.

See The Solution Submitted by Bractals    
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Some Thoughts attempt at the rest | Comment 2 of 3 |

a. The null set contains nothing, not even itself, so we're seeing {} anded with {null set}. The anded condition leads to {} or in other terminology, the letter phi. So the RHS has an extraneous set of braces, as phi already incorporates a set of braces, as it is a set. The RHS is basically {{}} and it should only be {}, or phi. This one is false.

b. phi is already a subset of {phi} as mentioned in my previous post. Therefore it adds no new element to {phi}. This one is also false.

c. One of the elements within the parentheses on the RHS is phi. One of its subsets is the null set (not the one that is shown as phi, a member, but the subset phi that's a subset of all sets). Since phi is a subset of the parenthesized set, it is a member of the set designated by the omega, and the membership claim is True.

d. While the null set is a subset of every set, the set consisting of the null set as an element is not. Therefore it is false that the set shown is a subset of any arbitrary set A.

More later, when I have more time.

  Posted by Charlie on 2014-04-19 11:01:51
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