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I am just beginning to learn sets and am in need of a little guidance. The problem I need help with is: $A = \{a, b, c\}$, $B = \{\{a\}, \{b\}, \{c\}\}$. I believe that $A$ is not a subset of $B$, and $B$ is not a subset of $A$. I think my main question is if what I believe is correct. Is $a = \{a\}$ ?

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  • $\begingroup$ Nope. it is not correct. $\endgroup$ – OmG Jan 10 '18 at 16:56
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I hate this, I hate this, I hate this....

... but it tends to work and is useful for people.

A set is a bag full of groceries. The elements of a set are the individual groceries inside the bag. The bag of groceries is a "thing" in an of itself and is not the groceries that make it.

And it the store you can buy, smaller bags of groceries and put them in your bag.

So $A = \{apple, banana, carrot\}$ is a bag with three things in it. They are apple, banana, carrot. But $B=\{\{apple\},\{banana\}, \{carrot\}\}$ is a bag with three things in it. The three things are a BAG with an apple in it, a BAG with a banana it it, and a BAG with a carrot in it.

An apple is a completely different thing that a bag with an apple in it. These are not at all the same.

Anyway I really hate that analogy as it implies the skin of the bag made of paper is itself an thing and it avoids the question of "Well, gee, if an apple is inside a bag, and the bag is inside a bigger bag, isn't the apple also inside the bigger bag just deeper?" The answer to that is a resounding no; a set is concept of the collection and although a collection may have a set as an element that does not imply the elements inside the set are in the collection.

An analogy I prefer is a library and you get to the reference desk and there is a reference book called "The Encyclopedia of Associations" which includes the Boy Scouts. And there is a reference book called "Books in Print" which includes "Razor Girl" by Carl Hiason. And there is a reference book called "Guide to Reference Books" which includes... "The Encyclopedia of Associations" and "Books in Print". But it does not include, for obvious reasons, either the Boy Scouts (which are not a reference book) nor "Razor Girl" (which is also not a reference book).

Of course, this analogy has it it problems too. Namely, these lists are references to things and not the things themselves. In sets the are the things themselves but being a collection things is different than the things being themselves rolling about without a bag.

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  • $\begingroup$ Thank you so much for this, it helped me understand the problem really well. $\endgroup$ – P. Patel Jan 11 '18 at 13:25
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Nope. it is not correct. As $a$ is a member of $A$ and $\{a\}$ is a subset of $A$. In other words, type of $a$ is an element, but the type of $\{a\}$ is a set.

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  • $\begingroup$ Does this mean that $A$ is a subset of $B$ too? Thank You in advance. $\endgroup$ – P. Patel Jan 10 '18 at 17:28
  • $\begingroup$ @P.Patel Definitely no. This means $A$ cannot be subset of $B$. $\endgroup$ – OmG Jan 10 '18 at 18:06
  • $\begingroup$ @OmG What if $a = \{b\}$ and $b=\{c\}$ and $c = \{a\}$? (which violates ZFC which proves A cannot be a subset of B-- but for different reasons). $\endgroup$ – fleablood Jan 10 '18 at 19:28
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Set $A$ contains 3 elements $a$,$b$,$c$. On the other hand set $B$ however, contains 3 sets that all contain 1 element each.
That means that every set in set $B$ (in your case $\{a\},\{b\},\{c\}$) is a subset of $A$.

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$A = \{a, b, c\}$, $B = \{\{a\}, \{b\}, \{c\}\}$. I believe that $A$ is not a subset of $B$, and $B$ is not a subset of $A$.

That's correct. $A$ and $B$ do not even share any objects, which is a requirement for non-empty subsets. $\{a,b\}$ is a subset of $A$. $\{\{a\},\{c\}\}$ is a subset of $B$.

I think my main question is if what I believe is correct. Is $a = \{a\}$ ?

No, that's not correct.

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