In no particular order:

x \in X = x is in X

X \ni x = X contains x (subtly different from the above)

x \notin X = x is not in X

Y \subseteq X = Y is a subset of or is equal to X

Y \subset X = Y is a proper subset of X i.e. not equal to X

f: X \to Y = the function f maps elements in X to elements in Y

f\;\circ \; g = function composition, do g first, and then do f

\exist x = there is some x

\forall x = for all x

\emptyset = the empty set

X \cap Y = X intersect Y i.e. the elements that X and Y share

X \cup Y = the union of X and Y i.e. the set that is all elements of X and all elements of Y in no particular order

X \times Y = Cartesian product i.e. the set whose elements are the ordered pair (x, y), x in X, y in Y

The following equalities may also be of some use;

X \cap \emptyset = \emptyset always

X \cup \emptyset = X always

X \cap X = X always

X \cup X = X always

Note: There are some standard AMS LaTex symbols not supported on this site

As I learn I am stealing from posts and/or adding facts (I think) to aid me remember (first revision - probably errors remain):

A "set" is a collection of LIKE elements. Indicated by listing the elements between { } and separating them by commas.
....Note these "elements" may be sets. E.g. {{a,b}, {b,c}} is a set and not the same as {a,b,c} as one has two, the other three elements and elements of one are "points" while the other's elements are sets.

A "Power set" (often written as \mathcal{P}(S) ) is constructed "on another set S" (actually from the elements of S) and called "the power set of S". It is the set of all possible sets that can be constructed (by only unions and/or intersections of the elements of set S) with the elements of set S.
....Note that power set always contain the null set as one of its elements. (By definition, I think, if there is only one element in the set and for two or more element sets, I think, the intersection of all elements of the set surely "constructs" the null set.)

Thus \mathcal{P}(S) = {{a,b}, {b,c}, {a,b,c}, {b}, emptyset} if the set S is {{a,b}, {b,c}}.
.... Note I am assuming here that none of the "elements" of set S can be divided or broken up during the "construction" of the power set. Specifically from this set, for example, one can not get either set {a,c} , or element c, or set {c} by the permited construction "tools" (unions and intersections).

A manifold is a topological space with certain additional properties. Thus must first define what a topological space is.

To be continued after "review and correction" by people who know.

....Note some or all of these "elements" may be sets. E.g. {a, {b,c}} is a set No, Billy, this makes no sense, as I was discussing with temur. A set is either a set of points, or a set of sets. You may not mix them up like that

....Note that power set always contain the null set as one of its elements. Well, of course - the null set ( empty set) is a subset of every set. By this:

Suppose that X,\;Y are sets, with X not empty (Y ambiguous).

If it is the case that X \cap Y = Y and X \cup Y = X, I will say that this defines Y as a subset of X (or is equal to it). So if Y = \emptyset it's easy to see that \emptyset \subset X.

(By definition, I think, if there is only one element in the set and for two or more element sets, I think, the intersection of all elements of the set surely "constructs" the null set.)Hmm, I don't understand this

set \{a,\;\{b,c\}\;, \{a,\;\{b,c\}\}, null set} is \mathcal{P}(S) Surely you made a typo here? Or maybe several? Whatever, this is not the powerset of anything!

No, Billy, this makes no sense, as I was discussing with temur. A set is either a set of points, or a set of sets. You may not mix them up like that

Well, in naïve set theory there's nothing with mixed sets - of course, you run into the Russell paradox, but barring that...

Well, in naïve set theory there's nothing with mixed sets - of course, you run into the Russell paradox, but barring that...I went to wiki to learn what "Russell's paradox" is. It seems to me to be the set theory version of following:

"This sentence is false."

or something quite like it, which is self contradictory.

Anyway, perhaps, if I were to use wiki I would learn the terminology and rules of set theory more easily, however, as I expressed to QuarkHead in PM, there are certain advantage great ignorance in a reasonably good brain can bring to an established field by noticing the tacit assumptions of the field, which years ago may have been explicit.

I may not continue to make and correct post 2's memory / set theory guide. Will look at wiki and decide later.

but barring that...Barring what, exactly, funkstar?

Well, of course - the null set ( empty set) is a subset of every set. ...
Surely you made a typo here? Or maybe several? Whatever, this is not the powerset of anything!You seem to be tacitly assuming that, for example a set can not be {5} where 5 is the fifth cardinal number. I.e. assuming that intersection of set {5,6} with {4,5} is not a set of one element.

If you reply that {5,6} is really {5,6,null set} etc for {4,5} then I ask about the intersetion of {3,4, null set} with {5,6,null set}. That surely is a set with single element so at least one single element set can exist. Why not {5}? Is it just "by definition" all sets contain the empty set? Then the empty set is not empty! (it is self containing!)

Or am I missunderstanding you?

My "typo" was probably just that I can not use Latex correctly and perhaps also related to my erroneous assumption that one could have mixed sets (elements not all of the same type)

I really do not see why not "mixed set" as both unions and intersections of their elements are well defined. E.g. if {a, {b,c}} were a set, then the union of its two elements is the set itself and the intersection is the null set. If the set were {a,b,{c,d}} then some unions are not the set itself nor the nullset. Mixed set {a,b,c,{c,d},{d,e}} is even more interesting when unions and intersections are formed. Want to give its power set? I think its power set contains {d}, but not d, for example as a single element set if they were permitted sets. Surely it contains two element set {d, null set}

Please comment on "one element set" and post 2, if more is still wrong there.

You seem to be tacitly assuming that, for example a set can not be {5} where 5 is the fifth cardinal number.

Billy---

I think a set is defined to have an element such that a + i = a, called the aditive identity.

Now I will wait for QuarkHead's scolding, because I am just guessing here and may well be wrong:)

Actually, in set theory it probably takes the form

X \cup \emptyset = X

I think a set is defined to have an element such that a + i = a, called the aditive identity You are thinking of algebraic groups? - sets do not necessarily include any operations, let alone identities under them.

You are thinking of algebraic groups? - sets do not necessarily include any operations, let alone identities under them.

Well, no---I'm not a mathematician so I don't know. Intuitively, however, it seems that sets should, at least tacitly (as Billy pointed out), include a null element. This would make the requirement

X \cup \emptyset = X

make sense.

I think it is not that a set is including a null (the empty set) element, but rather the existence of the empty set in set theory, when you study sets and their inter-relations.

You seem to be tacitly assuming that, for example a set can not be {5} where 5 is the fifth cardinal number.

There's nothing wrong with this; {5} is a set. So is {5,6}. Ony way to form a proper subsets is to remove elements from the set. Removing the element '6' from the set {5,6} yields the set {5}. Removing the element '5' from this set yields the null set.

Suppose a\in A. Then the set B that results from removing a from A can be written as

B=A\setminus\{a\}=\{x\in A:x\neq a\}

I guess I'm sort of losing the plot here. First, I don't think I ever denied the existence of the singleton set, nor would I ever. But I would insist that a and {a} are not the same thing. I offer two arguments;

\{\emptyset \} is a singleton set, a set with exactly one element. \emptyset is not a singleton, as it has no elements. Or if you prefer; \{a, b, c\} is a three-element set, whereas \{\{a,b,c\}\} is a singleton.

Second, I'm glad to see that Ben corrected himself; the notion of an "additive identity", and indeed the operation "addition" are not defined in set theory.

But to repeat myself: the following are all distinct sets \{a,b,c\},\;\{c,d,e\},\;\{x\},\; \emptyset. Unions, intersections etc are set theoretic binary operations, i.e. they operate on sets, so \{a,b,c\} \cup \{c,d,e\} = \{a,b,c,d,e\} makes sense, so does \{a\} \cup \{b\}=\{a,b\} but a \cup b doesn't.

... I would insist that a and {a} are not the same thing. I offer two arguments; ...No necessary, I think, as I believe no one is suggesting a is any set, not even the set {a}.

No one has responed to my arguments for "mixed" sets (IN POST 7) but those better versed in set theory will not accept the "mixed set" so for convenience I repeat my argument for it being reasonable to allow mixed set to be a set:

" ... I really do not see why not "mixed set" as both unions and intersections of their elements are well defined. E.g. if {a, {b,c}} were a set, then the union of its two elements is the set itself and the intersection is the null set.

If the set were {a,b,{c,d}} then some unions are not the set itself nor the nullset. Mixed set {a,b,c,{c,d},{d,e}} is even more interesting when unions and intersections are formed. Anyone want to give its power set? I think its power set contains {d}, but not d, for example, as a single element set. Surely it contains two element set {d, null set} ..."

Same as in post 7 except I have eliminated: "if they (single element sets) were permitted sets" from post 7 text as now all agree, I think {a} is a set of one element.

PS, based on D.H.'s post, Temur's post 12 comments and consistent logic, I tend to think that you (In middle section of post 3) are not correct in stating that "all sets contain the null set." I pointed out in prior post that if that were true then the null set also being called the "empty set" is nonsense as no set can be "empty" if all sets contain (implicitly) the null set. I think the correct POV is that the null set is the result of Either removing all elements from a set OR the set that results for the intersection of two sets that have no common element, period.

No necessary, I think, as I believe no one is suggesting a is any set, not even the set {a}.

You are doing this right here:

" ... I really do not see why not "mixed set" as both unions and intersections of their elements are well defined. E.g. if {a, {b,c}} were a set, then the union of its two elements is the set itself and the intersection is the null set.
'a' is not a set. You cannot take the union of a non-set ('a') and a set ({b,c}).


PS, based on D.H.'s post, Temur's post 12 comments and consistent logic, I tend to think that you (In middle section of post 3) are not correct in stating that "all sets contain the null set." I pointed out in prior post that if that were true then the null set also being called the "empty set" is nonsense as no set can be "empty" if all sets contain (implicitly) the null set.

You are conflating the concepts of containment and inclusion. An example of the distinction between the two terms: The set {5,6} contains 5 as an element. The set {5,6} includes {5} as one of its proper subsets. {5} and 5 are very different things. Containment and inclusion are distinct concepts.

Note to QuarkHead: Add inclusion to your glossary.
X\supseteq Y = X is a superset of or is equal to Y
X\supset Y = X is a superset of Y

This conflation of terminology is part of the problem you are having here. The set {5,6} does not contain the null set as the null set is not an element of the set. On the other hand, the set {5,6} does have the null set as a proper subset.

I think the correct POV is that the null set is the result of Either removing all elements from a set OR the set that results for the intersection of two sets that have no common element, period.

An immediate consequence of this first POV is that the null set is a subset of every set.

Barring what, exactly, funkstar?
Barring the Russel paradox, naïve set theory (as opposed to axiomatic set theory like ZF or NBG) is perfectly adequate to describe all these concepts. Which means that mixed set are quite allowable...

[Edit:] I'm not trying to knock the thread, btw. It's very nice work you're doing, QuarkHead.

No necessary, I think, as I believe no one is suggesting a is any set, not even the set {a}.Good, I'm glad to hear it! But then.........
" ... I really do not see why not "mixed set" as both unions and intersections of their elements are well defined. E.g. if {a, {b,c}} were a set, then the union of its two elements is the set itself and the intersection is the null set.See my problem? As I said in my last: union is a binary operation on sets. You may not union an element (that is not itself a set) with anything, you may only union sets!!

I tend to think that you (In middle section of post 3) are not correct in stating that "all sets contain the null set." I pointed out in prior post that if that were true then the null set also being called the "empty set" is nonsense as no set can be "empty" if all sets contain (implicitly) the null set. .Well, I confess I don't follow this at all. Each and every set contains itself, right? This is therefore true of the empty set. I do not see your point.

OK. Would you agree that any subset is contained in its superset? Would you also agree with the following definition of a subset:

A \subseteq B iff A \cup B = B and A \cap B = A?

Would you agree that \emptyset \cup B= B and \emptyset \cap B= \emptyset? whether or not B = \emptyset?

Simple substitution requires that \emptyset \subseteq B

If the answer to any of the above is "no", we are talking a different language.

It is perfectly true, as has been hinted at, that, in Set Theory, I may define the set B \setminus A to denote the set B without the subset A.
In set theory, this is called the complement of A in B, wherefore A \cap B \setminus A = \emptyset. Now try this by letting A = \emptyset (bet you get in a tangle!)

Well, I confess I don't follow this at all. Each and every set contains itself, right?

There is some overloading of terminology here that is hindering Billy's understanding. In the quoted text, you use the word "contains" to mean the symbol \supseteq. In the first post, your glossary use the word "contains" to mean the symbol \ni. Completely different concepts, containing as a subset and containing as an element, same English word. It might help to add superset to your terminology and explain how these two meanings of containment are quite different beasts.

Good point D H. Using the AMS LaTex guide I tried \superseteq for superset, but it was denied. Glad to see you found it! I agree they are different concepts; let me straight away make amends, but note this thread was originally intended as a glossary of terms, not a set theory tutorial.

Ah, looks like I can't edit the opening post. Damn!

Anyway, I'm going to the pub, so I guess this is the last sense you'll get out of me today!

Quark---

tell me what you want to change in your OP and I will edit it.

Good, I'm glad to hear it! But then.........
See my problem? As I said in my last: union is a binary operation on sets. You may not union an element (that is not itself a set) with anything, you may only union sets!! I am not disputing any of the facts, but that rule you just stated is arbitray also. I.e., except for it, why is not a\cup \{b,c}\ = {a,{b,c}} (Hope my stolen copy and modified of La tex worked.) By second edit: I now almost have it the { } around the b,c are strangely large I.e. I wanted something like: aU{b,c} = {a,{b,c}}.

Ain't "trial and error" grand?:D

Yes I agree with your logic / proof, but if "union" were generalized and "mixed sets" were not also arbitarily excluded from being sets, I do not see any problem with the above result of a union between the element a and the set {b,c} being (mixed) set {a,{b,c}} which is what I hoped my Latex above would also state.

BTW, hope you enjoyed vist to pub. I will be away from Sao Paulo for 3 days beginning in a few hours, but may have internet access. I have not "given up" yet, but think I will stop trying to learn Latex simultaneously.

OK, folks, I have three problems here. First is, I went to the pub (obvious consequence!)

Second, I sat on my spectacles, so I am really struggling here

Third, and most important; this thread is supposed to be about notation. If anyone wants to start a "set theory" tutorial, then please do. I had not intended this thread to be a tutorial in that sense.

Sorry I came across as grumpy last night - I was pissed off about my specs. Anyone is, of course free to post any set theory -related comment, ignore me.

Anyway, let me say a word or two about usage, as D H raised an interesting point.

If I write x \in X I mean something like this: given the set X, then x is an element in X. No harm is done if I use the negation:given the set X there are no elements in X - I have simply defined the empty set.

Now if I write X \ni x I mean something like: given the element x, then X is a set containing x. The negation - given the element x there is no set containing x is the Russell paradox, since this statement merely defines the set whose elements are members of no set.

Now consider A \subset B. (the following is easier to see for proper subsets). Here I mean that all elements in A are also elements in B, but there are elements in B that are not also elements in A.

But here I am under an obligation to give rather specific instructions on how to find the subset. This is not usually especially onerous - all I require is a "such that" or "for which" statement. So I am saying : B is the set of which A is a subset.

Note there is no particular consequence to the negation - B may well have no non-empty proper subsets.

If I write B \supset A, I am saying that given the set A then B is a superset of A. But note, the negation does not imply a paradox, merely that A is not a subset of any other set.

In short; the assertion X \ni x must be true, the assertion X \supset Y need not be; they have, as assertions, a different status.

Sorry, in my eagerness to "sit on" Billy T, I missed these points:Note to QuarkHead: Add inclusion to your glossary.
X \supseteq Y = X is a superset of or is equal to Y
X \supset Y= X is a superset of Y

Ben can you do this us please? (The placement in the list should be fairly obvious)

By the way, for proper subsets I much prefer X \subsetneq Y, but as you see it doesn't render here.

naïve set theory (as opposed to axiomatic set theory like ZF or NBG) is perfectly adequate to describe all these concepts. Which means that mixed set are quite allowable...You have the better of me here! By naive set theory do you mean a construction like 0 = {}, 1 = {{}} and so on? If so, it looks like you are right about "mixed sets". But I only know ZF, and even then, only what I need to know.

I have to say, that I have never encountered the usage practice you present here, QuarkHead. In my experience, the semantic content of X \ni x and x \in X is exactly the same: Set X contains the element x is equivalent to stating that element x is contained in set X. The same for the negation: As far as I know, x \notin X and X \not \ni x mean exactly the same thing.

You have the better of me here! By naive set theory do you mean a construction like 0 = {}, 1 = {{}} and so on?

No, in naïve set theory one would not worry about the representation of the integers or anything like that, but merely assume that "they are there", so to speak. And the problems of considering pretty much anything as being a set, and the consequent problems is, of course, what led to the more rigourous approaches such as ZF(C).

I merely meant that in order to learn the concepts presented here, the rigour used in axiomatic set theory is not necessary, and there's no problem with considering, say, \{a,\{a\}\} as a set, even though it is "mixed".

I have to say, that I have never encountered the usage practice you present here, QuarkHead. In my experience, the semantic content of X \ni x and x \in X is exactly the same: Set X contains the element x is equivalent to stating that element x is contained in set X. The same for the negation: As far as I know, x \notin X and X \not \ni x mean exactly the same thing.You perhaps know better than I do, as, like most, I absorbed set theory more-or-less by osmosis. See if you can agree with this:

Let x be an object in the universe \mathcal{U}_o of objects. I will say that some set X is well-defined iff x \in X or x \notin X.

Suppose now that X is well defined and that \forall x \in \mathcal{U}_o,\; x \notin X, then X = \emptyset.Right?

now suppose there is some set X in the universe \mathcal{U}_S of sets. If it is the case that there is some x \in \mathcal{U}_o where \forall X \in \mathcal{U}_S,\; X \not\ni x then this invites the wrath of Russell.

Hence my contention that x \notin X and X \not \ni x are logically inequivalent. No?

There is no absolute "universe" since it will lead to Russel's paradox. So we just construct sets starting from smaller ones. And x\in X and X\ni x are logically equivalent. They are only different in that when other symbols are present using one or the other you can make expressions shorter. For example, \exists X\ni x can be written as \exists X such that x\in X.

For example, \exists X\ni x can be written as \exists X such that x\in X.

What kind of backwards math are you doing?

;)

I've never seen statements written like this. I think a few of my professors would mark me off for using "ambiguous language" if I said \exists X\ni x.

Of course, there's nothing wrong with it.

There is no absolute "universe" since it will lead to Russel's paradox. Yes it would had I chosen to call \mathcal{U} the universal set, but I didn't; \mathcal{U} was intended merely to denote " all objects/sets you you possibly imagine".

So we just construct sets starting from smaller ones.And at what point in this process do you feel Bertie's breath on your neck?

And x\in X and X\ni x are logically equivalent. They are only different in that when other symbols are present using one or the other you can make expressions shorter. For example, \exists X\ni x can be written as \exists X such that x\in X.
temur, you miss the point; suppose I write, given the set X, that x \notin X\; \forall x. How would you describe X? \emptyset perhaps?

Suppose I now write, given the point x that X \not \ni x\; \forall X. What would you say to that? Russell paradox, maybe?

This, it seems to me, is a logical inequivalence

If I write x \in X I mean something like this: given the set X, then x is an element in X.

Now if I write X \ni x I mean something like: given the element x, then X is a set containing x.
Both statements are equivalent (unless you think you can find a set X and and element x such that x \in X and X \not\ni x, or vice-versa). In your definition of x \in X you've chosen to put emphasis on the fact that x is generally not the only element contained in X, and for X \ni x you've stressed that X may not be the only set that contains x, but both statements apply in general.
No harm is done if I use the negation:given the set X there are no elements in X - I have simply defined the empty set.
The negation of "x is an element of the set X" is "x is not an element of the set X".

"There are no elements in X" is the negation of "There exists at least one element in X".
The negation - given the element x there is no set containing x is the Russell paradox, since this statement merely defines the set whose elements are members of no set.
The negation of "X is a set containing the element x" is "X is not a set containing the element x".

"There is no set which contains x" is the negation of "There exists at least one set which contains x".
If it is the case that there is some x \in \mathcal{U}_o where \forall X \in \mathcal{U}_S,\; X \not\ni x then this invites the wrath of Russell.

How would this statement be different if you substituted X \not\ni x for x \not\in X?
I can define S_x := \{ x \} for any given x \in \mathcal{U}_o, so no x will ever satisfy the condition above.

given the set X, that x \notin X\; \forall x. How would you describe X?

Suppose I now write, given the point x that X \not \ni x\; \forall X.
How about:
X such that X \not\ni x\; \forall x

and
x such that x \not\in X\; \forall X

?

I've never seen statements written like this. I think a few of my professors would mark me off for using "ambiguous language" if I said \exists X\ni x.

I agree this is a bit strange, but I often saw professors writing like this. For example, I think X\ni x\mapsto y\in Y is standard.

edit: \mapsto shows a different symbol (\rightarrow) here.

Yes it would had I chosen to call \mathcal{U} the universal set, but I didn't; \mathcal{U} was intended merely to denote " all objects/sets you you possibly imagine".


Yes, I agree. It is often clear from the context what "universe" you are in. I think you are thinking in terms of operations and the order in which they are performed; but sometimes it is useful to think of mathematics just as symbols written on a piece of paper or blackboard. Actually there is no "time" in mathematics, things just exist.

http://en.wikipedia.org/wiki/Zermelo-Fraenkel_set_theory#Axioms

Well, you all seem intent on challanging a statement I never made!

Look. No sane person would deny that x \in X \Rightarrow X \ni x, and I am sane (I think).

Likewise, clearly x \notin X \Rightarrow X \not \ni x.

It's possible I am being unorthodox, and I accept that it is largely a matter of notational aesthetics. It is simply, as pryck noted, where one places the emphasis. But surely all would agree with the following:

If I write that, for some fixed set X, that x \in X, without an explicit quantifier, then I mean that x is a typical (generic) element in X; anything I say about x can be taken to be true of all elements in X, unless I say otherwise.

If I write that, for some fixed element x, X \ni x, it is not the case that whatever I say about X can be taken to be true of all sets containing the element x.

All this is so, obviously, because X is a set, and x is an element.

As I say, it's merely notational aesthetics, and therefore scarcely something to be dogmatic about. Time to move on, I say!

Let's move on!

In no particular order:

x \in X = x is in X

X \ni x = X contains x (subtly different from the above)

x \notin X = x is not in X

Y \subseteq X = Y is a subset of or is equal to X

Y \subset X = Y is a proper subset of X i.e. not equal to X

f: X \to Y = the function f maps elements in X to elements in Y

f\;\circ \; g = function composition, do g first, and then do f

\exist x = there is some x

\forall x = for all x

\emptyset = the empty set

X \cap Y = X intersect Y i.e. the elements that X and Y share

X \cup Y = the union of X and Y i.e. the set that is all elements of X and all elements of Y in no particular order

X \times Y = Cartesian product i.e. the set whose elements are the ordered pair (x, y), x in X, y in Y

The following equalities may also be of some use;

X \cap \emptyset = \emptyset always

X \cup \emptyset = X always

X \cap X = X always

X \cup X = X always

Note: There are some standard AMS LaTex symbols not supported on this site

Excellent work... yet again... this place does indeed surprise me sometimes. :bawl:

:D

Now... [please] leave him alone. His work is mathematical perfect.

\exist x [/tex] = there is some x

How about adding this:

\exist !x = there exists some unique x"

Somebody told me that \exists x should be read as "for some x" not "there exist..".

Somebody told me that \exists x should be read as "for some x" not "there exist..".

They both mean the same thing. But the backwards E is there to symbolize the e in exist. Similarly, the upside down A is the a in "for all."

I find it easier to deal with "there exists."

Yes, they do mean the same thing. "Some dog is black" \equiv "a black dog exits"

I think the OP should try to define the logical symbols via truth tables. That way semantic confusions are eliminated.