Problem in understanding set theory

I am stuck with a small question. It is below,

Let $A$ be a set and let $B = \{A,\{A\}\}$. Then, since $A$ and $\{A\}$ are elements of $B$, we have $A \in B$ and $\{A\} \in B$. It follows that $\{A\}\subseteq B$ and $\{\{A\}\} \subseteq B$. However, it is not true that $A\subseteq B$.$\hspace{302pt}\blacksquare$

I have to ask two questions from this text,

1. What is the difference between $A$ and $\{A\}$?
2. As $A$ is contained by $B$ so $A$ belongs to $B$ should be true but it is not. Why is that?

Thanks.

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It would be necessary for you to specify under which set theory you are working. In my answer below I assumed you were dealing with Zermelo-Fraenkel (ZF). – lentic catachresis Jul 8 '11 at 4:57
I always like saying, "If $A$ is a bunch of oranges in a bag, then ${A}$ is a bunch of oranges in a bag sitting inside another bag. – james Jul 8 '11 at 15:14
@james: I think you lost some braces in your comment. To display them, you need a backslash in front of them. So to get $\{$, you put \{ between dollar signs. – Ross Millikan Jul 8 '11 at 16:29

(1) A is a set with certain elements, and {A} is a set with only one specific element, and that element is itself a set: the set A. In set theory we not only regard the elements of a set as mathematical objects, but the sets themselves as mathematical objects, and we distinguish between the two. For example, the numbers 1, 2, and 3 are all numbers whereas the set {1,2,3} is not a number but a set of numbers. Furthermore, the set {{1,2,3}} is not a set of numbers (as {1,2,3} isn't a number) but a set containing a set of numbers.

(2) A is an element of B, that is $A \in B$. However, that doesn't mean any of the elements of $A$ are actually straight-up elements of $B$. For example, say A is the set {1,2,3}. Then B = {{1,2,3},{{1,2,3}}}. Neither {1,2,3} nor {{1,2,3}} are the numbers 1, 2, or 3, nor are they numbers at all (they are sets), so we can't say any of 1, 2, or 3 are in the set B.

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Well, in your answer to 2), you're not really proving that it is never true that $A\subseteq B$, you're just giving an example where it is not true... – lentic catachresis Jul 8 '11 at 15:53
Indeed. What someone just starting to understand set theory needs is more of an illustration of the mechanics of inclusion and the distinction between the levels of hierarchy of sets rather than a formal proof. Hence the explanation-by-example. The concept generalizes to any situation fairly easily once you "get it" I think. – anon Jul 8 '11 at 21:52
1. $A$ is a given set, and $\{A\}$ is the set which has only one element, namely $A$.

2. Be careful, it is true that $A\in B$. That is, the set $A$ is an element of the set $B$. It is not true, though, that $A\subseteq B$. Why? Well, by definition: $$A\subseteq B \iff (x\in A \Rightarrow x\in B)$$ This is not true, because there is an element of $A$ that is not an element of $B$. In fact, none of the elements of $A$ can be elements of $B$, for the elements of $B$ are only $A$ and $\{A\}$, but $A\notin A$ and $\{A\}\notin A$. See Axiom of regularity

Indeed, it is not true that $A\in A$ because "no set is an element of itself" as explained in the Wikipedia link (in Zermelo-Fraenkel theory which is a well-founded theory), and it is not true that $\{A\}\in A$, for this would imply that $A\in \{A\}\in A \in \{A\}$... forming an infinite descending sequence of sets, which is also impossible by regularity.

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Just to illustrate the point everyone is making, if $A$ is The Adventures of Tom Sawyer, then $A$ is a book, but $\lbrace A\rbrace$ is a (very small) library. – Gerry Myerson Jul 8 '11 at 4:42
@Gerry: nicely put. I was going to comment it in this form: "A bag is not the same as a bag containing a bag". – lentic catachresis Jul 8 '11 at 4:45

$A$ is the set $A$ itself, but $\{A\}$ is a singleton set with just one element, namely $A$. Yes, sets themselves can be elements of other sets.

You don't know that $A$ is contained by $B$, you only know that $\{A\}$ is contained by $B$, $\{A\}\subseteq B$. This is because the only element $A$ of $\{A\}$ is also an element of $B$. This is the definition of what it means to be a subset.

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1. $A$ is a set containing elements (which are not explicited here, by the way) and $\{ A \}$ is a set containing one element and this element is the set $A$. The elements of $A$ are not elements of $\{ A \}$.

The relation $A \in B$ means that $A$ is an element of $B$. The relation $A \subseteq B$ means that every element of $A$ is an element of $B$. Since the only elements of $B$ are $A$ and $\{ A \}$, unless $A$ has only A or $\{ A \}$ as an element (which would make a recursive definition which is kind of weird... I don't know if this is somehow possible but I think it's not allowed under some axioms, to see how weird it could be, you would have $A = \{ A, \{A\} \}$, $A = \{A \}$ or $A = \{ \{ A \} \}$, and to be honest usually you don't consider those cases), you can't say that $A \subseteq B$.

Hope that helps,

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When you downvote, at least take the time to explain why. Putting a downvote without any comment is mean and purpose-less. – Patrick Da Silva Jul 9 '11 at 22:18