Please feel free not to read this its just a prelude: Ok I am sorry for making this question since as far as I can tell the level of the other questions here is higher by far compared to mine, but I am an adult that wants to learn maths from scratch (well I have passed my high school maths when i was younger and have passed some basic calculus etc exams when i was in university but my knowledge consist of fragments and is not well structured even at that level described above I am able to solve the problems I can solve mainly because I follow predefined steps -having some notion of what im doing but still feeling lost in the details ) But now I would like to fill the gabs and dont care how far/deep I could get in the field of mathematics as long as what I currently achieve to learn is solid and undisputable. I also cant afford a tutor so I bother you guys :) if this isnt the forum for me please link me to an other forum thanks and sorry again.

Actual Question: So I have "worked" with sets and I have a very general knowledge about them but since I decided to start from scratch again to fill the gaps I began with this book: http://www.people.vcu.edu/~rhammack/BookOfProof/

And I got stuck in the first pages :P More specifically on page 4 it states

"It is even possible for a set to have other sets as elements. Consider E={1,{1,2},{2,4}}. which has three elements: the number 1, the set{2,3} and the set {2,4} this 1εE and {2,3}εΕ and {2,4}εΕ. but note 2 is not in E,3 is not in E and 4 is not in E."

So thats what I am not sure of... if {2,3} is a subset n1 consisting of 2 and 3 of the set E and {2,4} a subset n2 of the set E so that E=(1,n1,n2) then why the elements 2,3,4 are NOT in E (but are elements of n1 and n2 sets which are elements/subsets of E)?

Is n1 just a pair of coordinates? if yes how to distinguish that? is there an other notation that does that better? or you have to distinguish that using the context?

For me (meaning my current knowledge) the only to accept that 2,3,4 are not in E is only if n1,n2 are pairs of coordinates.

  • $\begingroup$ The reason is that in general, an element of an element is not an element of the bigger set, except for special sets known as ordinals. A simple metaphor: is it always true that! My friends' friends are my friends? $\endgroup$ – Bernard Aug 22 '15 at 19:16
  • $\begingroup$ Oh I see now so if n1= {2,3},n2={2,4} are functions lets say n1 f(x)=x+1 and n2 f(x)=x+3 Then if E= {1,{2,3},{2,4} } or E= {1,n1,n2} Then f(x)=x+x,xεΕ Has values of x={1,3,4,5,7} and will result to y= {2,6,8,10,14} is that right? $\endgroup$ – John Aug 22 '15 at 19:30

$2$ and $3$ are elements of the set {2,3} , but not of the set $E$.

A set is different to the elements of the set. And {2,3} is not a subset of $E$ because it contains elements (namely $2$ and $3$) which do not belong to $E$.

In short : If a set $A$ contains a set $B$ as an element, the elements of $B$ are in general no elements of $A$. Do not mix a set with its elements.

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  • $\begingroup$ Could you please elaborate a little more? I understand the statement (what you say that cant be) but i dont understand why it cant be. So in other words f(x)=x, xεE does not have f(3) and f(4) images? And if so when do {2,3} and {2,3} starting to "come into play" as elements? Are {2,3},{2,3} two pairs of coordinates? ? $\endgroup$ – John Aug 22 '15 at 19:21
  • $\begingroup$ Elements of a set need not be real numbers. So, they need not have to do anything with functions or coordinates. For example, the set {a,b,c} has letters as elements. $\endgroup$ – Peter Aug 22 '15 at 19:23
  • $\begingroup$ Of course, the concept of sets being elements of a set, is strange at first sight and not without problems ( See : en.wikipedia.org/wiki/Russell's_paradox ). But, it is useful to allow sets as elements of a set (See : en.wikipedia.org/wiki/Power_set ) $\endgroup$ – Peter Aug 22 '15 at 19:27

take this simple example to understand the concept Let $E$ be the set containing "James, Peter" we all know that James has a hand but "the hand" is not obviously an element of $E$.

to convince you more you know that the first axiom of set theory (ZF) is the axiom of extensionality which say that "A set is determined uniquely by its members" if we consider that in your example $E$ contient 2, and 3, and 1 only we will have $E=\{1,2,3\}$ which obviously false.

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  • $\begingroup$ so if n1= {2,3},n2={2,4} are functions lets say n1 f(x)=x+1 and n2 f(x)=x+3 Then if E= {1,{2,3},{2,4} } or E= {1,n1,n2} Then f(x)=x+x,xεΕ Has values of x={1,3,4,5,7} and will result to y= {2,6,8,10,14} is that right? $\endgroup$ – John Aug 22 '15 at 19:34
  • $\begingroup$ No, I don't understand you question right but if you take a set of numbers all element of this set will be a numbers, if your set is a functions all element of your set will be a functions not a numbers and vice versa $\endgroup$ – Hamza Aug 22 '15 at 19:38
  • $\begingroup$ in your example your set in mixed by a number which is 1 and two sets which are $\{2,3\}$ and $\{3,4\}$ and so they will not contient other numbers then 1. $\endgroup$ – Hamza Aug 22 '15 at 19:40
  • $\begingroup$ Well I am trying to reason as to why what "you" say is true (not that I doubt it i just cant reason with it) and to try to reason with it I made this thought: So if a set E consisted of {1,n1,n2} were n1 and n2 are functions like n1(x)=x and n2(x)= x with domains Dn1={2,3} and Dn2={2,4} And if there is an other function f(x)=x were i use the set E as a domain for that function so Df={Ε/for every x in E} So the function f would result in 1.2,4,2,3 using each x respectively, if that statement is true then I think I got it if its false then I dont know what{2,3} and{2,4} are in my main question $\endgroup$ – John Aug 22 '15 at 20:07
  • $\begingroup$ yes they are entities of an element of E but what is the use of them practically how do they "come in play" as entities of an element? $\endgroup$ – John Aug 22 '15 at 20:08

This is a typical problem on understanding sets.

The point is to distinguish well from elements of a set a subsets of the same set.

The elements have to be thinked as single entities, also if they are collection. So, in your case, $\{1,2\}$ and $\{2,3\}$ are single entities (that are sets formed of two numbers). But the two numbers that are in them are not elements of the set $E$ because are not single entities that form such set.

A subset of a set is a set of his elements, this means that it is an entity different from the elements of the set, being a set ( i.e. a collection) of elements.

So, $\{2,3\}$ is an element af $E$ and is not a subset of $E$ but $\{\{2,3\}\}$ is a subset of $E$ formed of the only one element $\{2,3\}$ (a set, but not a subset of $E$).

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  • $\begingroup$ so if Dn1= {2,3},Dn2={2,4} and n1 and n2 are functions lets say n1 f(x)=x+1 and n2 f(x)=x+3 Then if E= {1,{2,3},{2,4} } or E= {1,n1,n2} Then f(x)=x+x,xεΕ Has values of x={1,3,4,5,7} and will result to y= {2,6,8,10,14} is that right? $\endgroup$ – John Aug 22 '15 at 19:46
  • $\begingroup$ No! the function has to be defined on the set $E$ and have to be values on ana other set ( waht??). And the function is well defined only if you specifies its value for any element of $E$. Aniway, the first step is to well undesrtand what is a set, than you can think to fuctions. $\endgroup$ – Emilio Novati Aug 22 '15 at 20:28

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