# (SR) Glossary of Set Theoretic Symbols

##### Remedial Math Student
Valued Senior Member
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.

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....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.

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.

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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

$$X \cup \emptyset = X$$

ben said:
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.

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... 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.

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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.

$$X\supseteq Y$$ = 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.

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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.

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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.