Discrete Math: Subsets and Venn Diagrams Explanation

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The discussion centers on the relationship between set A, which is defined as the set of all integers, and set B, which contains A and the set containing A. It clarifies that while A is an element of B, it is not a subset of B because not all elements of A are included in B. The confusion arises from comparing individual elements of A with the sets themselves, leading to the realization that a set is a subset of another only if all its elements are contained within it. The conclusion is that {A} and {{A}} are subsets of B, while A and {{{A}}} are not. This highlights the distinction between elements and subsets in set theory.
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Homework Statement



Let their be a set A, and let B be the set: {A, {A}} (the set containing the elements A and the
set that contains element A)

As you know, A is an element of B and {A} is also an element of B.

Also, {A} is a subset of B and {{A}} is also a subset of B.

However, A is not a subset of B

Homework Equations



[URL]http://65.98.41.146/~carlodm/phys/123.png[/URL]


The Attempt at a Solution



See my drawing above. I created a Venn Diagram to deduce the logic with no clear results. In the first diagram to the left, I can clearly see that the element A is a subset of B ... yet.. it is not? Can someone explain?
 
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I don't think Venn diagrams are the way to go. Let's do an example. Let A be the set of integers. We know that B has two elements: A and {A}. But A has 3 as an element. 3 is neither the set of integers or the set containing the set of integers, so 3 is not in B.

In the venn diagram you are getting A as an object and A as a set containing other objects confused
 
My question:

Why are we comparing set A (the set of all integers) on an element-by-element basis? I agree that "3" is an element of A such that A = { ...,-1,0,1,0,1,2,3...}.

I agree that "3" is neither the set of integers A = { ...,-1,0,1,0,1,2,3...} nor the set containing the set of integers {A} = {{ ...,-1,0,1,0,1,2,3...}} .

I fail to see how individual elements of A are pertinent to the problem. If A is a subset of B, every element of A is in B.

----------------------------------------------------- Light Bulb in My Head

After writing the last line of my argument, I realized that simple truth:

"A is a subset of B iff every element in A is in B"

Given your argument, "3" is an element of A yet 3 is neither the set of integers NOR the set containing the set of integers.

Eureka moment. Thanks!
 
Going a mile further...

{A} is a subset of B in this example because...

Every element in {A} is in B.

The only element of the set of sets is A. A is the set of all integers. {A} therefore is the subset of B.

------------------------------------------------------------
{{A}} is a subset of B since:

Every element in {{A}} is in B.

The only element in the set of sets of sets is A. The sets of sets of A, {{A}} is an element of B.

That was correct, no?

{{{A}}} is not a subset of B... well because {{A}} is not in B?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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