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This is a total noob question.

I am reading Naive Set Theory by Paul R. Halmos, and I'm having difficulty to understand something which seems to be trivial.

In the first chapter he writes:

If $x$ belongs to $A$ ($x$ is an element of $A$, $x$ is contained in $A$), we shall write

$x~\epsilon~A$

I understand this.

Then, he write:

If $A$ and $B$ are sets and if every element of $A$ is an element of $B$, we say that $A$ is a subset of $B$, or $B$ includes $A$, and we write:

$A \subset B$

I understand this too.

Then he says:

The working of the definition implies that each set must be considered to be included in itself ($A \subset A$); this fact is described by saying that set inclusion is reflexive.

I understand this too.

But then:

Observe that belonging ($\epsilon$) and inclusion ($\subset$) are conceptually very different things indeed. One important difference has already manifested itself above: inclusion is always reflexive, whereas it is not at all clear that belonging is every reflexive. That is: $A \subset A$ is always true; is $A~\epsilon~A$ ever true? It is certainly not true of any reasonable set that anyone has every seen.

And this is where I don't think I understand anything. There is not more elaboration on this point in the text.

I tried to skip this but it seems it is quite fundamental for understanding what follows in the book.

Could someone explain what is meant here?

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2  
In the history of these things, "part of" (AKA subset) and "member of" took surprisingly long to disentangle. –  André Nicolas Apr 13 '12 at 14:38
    
@André: Speaking from a modern point of view, anything longer than ten minutes after coming up with the two notions would be surprisingly long. Do you have some reference for the history of these two? –  Asaf Karagila Apr 13 '12 at 14:47
    
@AsafKaragila: Can't think of references right now. But there is large literature. Much of Greek mathematics, in Aristotelian tradition, is built on "part of." This applies even to number theory. –  André Nicolas Apr 13 '12 at 15:19
    
@André: Oh, I thought you meant in modern times. Yeah, in that aspect I certainly agree. –  Asaf Karagila Apr 13 '12 at 15:42

3 Answers 3

up vote 3 down vote accepted

Whenever you come across something like this and it trips you up, you might want to look at particular examples. For instance, consider {4}. {4} $\subset$ {4}, but {4} $\epsilon$ {4} is false, since the only member of {4} is 4, not {4}. It may have tripped you up that "includes" and "contains" in everyday language usually qualify as synonyms. They don't here, and the terms get defined by the definitions for $\epsilon$ and $\subset$. You might want to prove that A $\subset$ A for any set A as it can get proven in a line or two.

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@AsafKaragila I didn't even notice that difference! –  Doug Spoonwood Apr 13 '12 at 13:40
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thank you, I think this made everything clear. It was very well explained. –  Komrade P. Apr 13 '12 at 20:02

The point is that every set is a subset of itself, namely $A\subseteq A$ - always.

However $\in$ does not have this property, for example $\varnothing$ has no elements, in particular $\varnothing\notin\varnothing$. However $\varnothing\subseteq\varnothing$.

To make matters worse, $\varnothing\in\{\varnothing\}$ as well $\varnothing\subseteq\{\varnothing\}$. However $\varnothing\neq\{\varnothing\}$ since the empty set has no elements and the singleton $\{\varnothing\}$ has an element.

In the axiomatic approach to set theory, the commonly used axioms of ZF dictate that $A\notin A$ for every set $A$. This is a result of something called the axiom of regularity, or axiom of foundation. However there are useful instances of non-well founded set theory in which some sets have the form $x=\{x\}$. For more information on that: When is $x=\{ x\}$?

Regardless to that, it is always the case that $A\subseteq A$, and always the case that for some $B$ we have $B\notin B$.

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Though undoubtedly $\subseteq$ works out as a better notation in general, the OP didn't use it here. So, it might work better to just use $\subset$ as the OP did. Or perhaps it's better to just say that you use $\subseteq$ instead of $\subset$, and maybe explain why. –  Doug Spoonwood Apr 13 '12 at 13:36
    
@DougSpoonwood, I used the notation verbatim from the text, including $\epsilon$ instead of $\in$ to give faithful representation of the original typography. –  Komrade P. Apr 13 '12 at 13:49
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@Komrade P. It is common nowadays to use $\in$ for membership and I personally prefer $\subseteq$ to avoid ambiguities (some authors use $\subset$ for proper inclusion, that is included but not equal). –  Asaf Karagila Apr 13 '12 at 13:57

If something belongs to set then it means thats it is an element of that set as a whole but if a set is a subset of another set then it means all the elements of that set belong to the set to which that set is a subset.

Ex: Lets take two sets A={1,2,3} & B={x:x is a natural number and x<5} Here, clearly evey element of set A is an element of set B hence we can say A is a subset of B but we can 't say A belongs to B as set A as a whole is not an element of set B.

Ex 2: A={1,2,3} & B={{1,2,3},4,5} Here set A is an element of set B itself. Hence we can say that A belongs to B but here a is not a subset of B as any individual element of A won 't be an element of set B.

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