[QUOTE=science_man_88;248993][B][U]Partition[/U][/B] :a breakdown of a set down into disjoint subsets.
[B][U]disjoint[/U][/B]: of or pertaining to sets having only [TEX]\empty[/TEX] in common.
[B][U]intersection[/U][/B]: the elements that 2 sets have in common.
[B][U]union[/U][/B]: a combining of 2 sets into a larger set containing no duplicates.
[B][U]n-tuple[/U][/B]: a ordered group containing a element from n not necessarily disjoint sets.
[B][U]collection[/U][/B]: a grouping of sets.
[B][U]pairwise disjoint[/U][/B]: pertaining to a collection having no combination of 2 sets such that there intersection contains anything but [TEX]\empty[/TEX]
[B][U]Cartesian Product[/U][/B]: the set containing n-tuples to show all combinations withing the elements of n, not necessarily disjoint sets.
[B][U]Binary relation[/U][/B]: a relation connecting 2 things.
[B][U]equivalence relation[/U][/B]: a binary relation that is reflexive, symmetric, and transitive.

Am I close to correct? found an error I made in these defintions looking back on the thread I think.[/QUOTE]

are there any other terms I should define ? I know equivalence classes are next in the text preliminaries some how I think they might screw me up as much as equivalence relations mainly because from the text they are one.

[QUOTE=science_man_88;249252]are there any other terms I should define ? I know equivalence classes are next in the text preliminaries some how I think they might screw me up as much as equivalence relations mainly because from the text they are one.[/QUOTE]

I'd be more interested in seeing you use these than having you add to the list.

one way to partition a chess board is white squares and black squares. that is a chess board is a collection of the sets W = white squares and B = black squares.
the chess board is also a Cartesian product of the sets (A,B,C,D,E,F,G,H) and (1,2,3,4,5,6,7,8). The main binary relations are "is black","is white" ,"is the same color as","is a different color than", "is in the same row as","is in a different row than", "is in the same column as","is in a different column than". lets take "is the same color as" as an example:

by definition a square is the same color as itself so by definition it is reflexive.

let us use 2 squares x and y if these squares have the same color as each other we can change the order and that relation (actually an equals relation) and it stays the same x=y y=x this is true so the relation is symmetric.

let us now set x and y equal as though they're the same square. since we know that the squares are the same color any relation to the last square chosen will be the same for both. so letting x=y we get x~x and x~z implies x~z the first part is trivially true by definition. the next part is assumed true the last part is the same as the second and hence taking the first part off the relation becomes trivially true since we can apply this to the first of the 2 squares of the same color we see that x~y and y~z implies x~z is true proving transitivity.

not a proof I know.

I forgot the terms to use so I peeked which I shouldn't have to, you win.

[QUOTE=science_man_88;249298]one way to partition a chess board is white squares and black squares. that is a chess board is a collection of the sets W = white squares and B = black squares.
the chess board is also a Cartesian product of the sets (A,B,C,D,E,F,G,H) and (1,2,3,4,5,6,7,8).[/QUOTE]

Good.

[QUOTE=science_man_88;249298]The main binary relations are "is black","is white" ,"is the same color as","is a different color than", "is in the same row as","is in a different row than", "is in the same column as","is in a different column than".[/QUOTE]

Usually I'd describe "is black" and "is white" as unary relations: I would say "x is black", not "x is black y". There are ways to define it as a binary relation, but it's not obvious which one is 'right'.

[QUOTE=CRGreathouse;249305]Good.



Usually I'd describe "is black" and "is white" as unary relations: I would say "x is black", not "x is black y". There are ways to define it as a binary relation, but it's not obvious which one is 'right'.[/QUOTE]

yeah I'm not thinking my brain shuts down sometimes.

[QUOTE=CRGreathouse;247576]Spot on, if you're describing the Cartesian product.

Now relations are (formally) described as a subset of the Cartesian product of the sets. So let's imagine we take just the white squares of the chessboard as our relation ~. Then since a1 is black, a ~ 1 is false. b1 is white, so b ~ 1 is true, and so forth.[/QUOTE]

think this is what i might have thought about.

[QUOTE=Mr. P-1;249146]Try to complete this sentence:

A partition on S is a collection of ______ of S, such that ...[/QUOTE]

[QUOTE=science_man_88;249150]1) Pairwise disjoint subsets of S, would be my best guess.[/QUOTE]

Close. A partition on S is a collection of [i]non-empty[/i] pairwise disjoint subsets of S such that every element of S is a member of one of the elements of the partition.

To express that last part in other words:

Let S be a set and let P be a partition on S. Then for every s[TEX]\in[/TEX]S there exists a p[TEX]\in[/TEX]P with s[TEX]\in[/TEX]p.

[i]Pay attention[/i]: it is common in mathematics to use the same letter in different cases to refer to different (but related) things. In the above s and S are not the same, nor are p and P.

Question: Could there be more than one p[TEX]\in[/TEX]P for each s[TEX]\in[/TEX]S? If so, give an example. If not, explain why not.

I don't think this has been spelled out fully: In set theory, the words "member" and "element" are completely synonymous. You can use them interchangeably. In the above, I used both to avoid the verbal monotony of saying "...is a member of one of the members of...".

[QUOTE]dearly noted on the word group I just didn't see set as a possibility but I should have though of the fact that the sets are sub[B]set[/B]s of the collection.[/QUOTE]

No! No! No. In a collection, the sets are [B]elements[/B] of the collection. They are not generally subsets of the collection.

For example, let A and B be sets, and let C be the collection {A, B}

The elements of C are A and B (which are sets.)
The subsets of C are [TEX]\empty[/TEX], {A}, {B}, and C itset.

The set A is not the same as the collection {A}. A might have any number of members. {A} has exactly one: the set A.

Question: Let S be a set. Is it possible that an element s[TEX]\in[/TEX]S also be a subset of S? If so, give an example. If not, explain why not?

[QUOTE=Mr. P-1;249933]Close. A partition on S is a collection of [i]non-empty[/i] pairwise disjoint subsets of S such that every element of S is a member of one of the elements of the partition.

To express that last part in other words:

Let S be a set and let P be a partition on S. Then for every s[TEX]\in[/TEX]S there exists a p[TEX]\in[/TEX]P with s[TEX]\in[/TEX]p.

[i]Pay attention[/i]: it is common in mathematics to use the same letter in different cases to refer to different (but related) things. In the above s and S are not the same, nor are p and P.

Question: Could there be more than one p[TEX]\in[/TEX]P for each s[TEX]\in[/TEX]S? If so, give an example. If not, explain why not.
[COLOR="Red"]No, because the term pairwise disjoint implies the element [TEX]s\in S [/TEX] not be part of more than one element [TEX]p\in P[/TEX] as the partition is like a collection of subsets . and pairwise disjoint implies they don't overlap.[/COLOR]


I don't think this has been spelled out fully: In set theory, the words "member" and "element" are completely synonymous. You can use them interchangeably. In the above, I used both to avoid the verbal monotony of saying "...is a member of one of the members of...".



No! No! No. In a collection, the sets are [B]elements[/B] of the collection. They are not generally subsets of the collection.

For example, let A and B be sets, and let C be the collection {A, B}

The elements of C are A and B (which are sets.)
The subsets of C are [TEX]\empty[/TEX], {A}, {B}, and C itset.

The set A is not the same as the collection {A}. A might have any number of members. {A} has exactly one: the set A.

Question: Let S be a set. Is it possible that an element s[TEX]\in[/TEX]S also be a subset of S? If so, give an example. If not, explain why not?

[COLOR="Red"]unless by definition the word set implies more than one element then yes it is possible that a element [TEX]s\in S[/TEX] could also be a subset. (though a rather small one)[/COLOR]
[/QUOTE]

this what you wanted ?

[QUOTE=davar55;249185]@P-1: For me at least please define both as you see the difference.[/QUOTE]

Intuitively, the [b]completion[/b] of a metric space is that space augmented with additional points so that every sequence that "ought to" converge actually does converge. Sequences which "ought to" converge are called Cauchy sequences.

For example, define the sequences a[sub]k[/sub], b[sub]k[/sub], and c[sub]k[/sub] within the rational numbers [b]Q[/b] as follows.

a[sub]0[/sub] = 1; b[sub]0[/sub] = 2.

c[sub]k[/sub] = (a[sub]k-1[/sub] + b[sub]k[/sub])/2 for k=[TEX]\ge[/TEX]1

a[sub]k[/sub] = c[sub]k[/sub] if c[sub]k[/sub][sup]2[/sup] < 2 else a[sub]k[/sub] = a[sub]k-1[/sub].

b[sub]k[/sub] = c[sub]k[/sub] if c[sub]k[/sub][sup]2[/sup] > 2 else b[sub]k[/sub] = b[sub]k-1[/sub].

a[sub]k[/sub], b[sub]k[/sub], and c[sub]k[/sub] are well-defined sequences of rational numbers which "ought to" converge to a limit whose square is 2. But no such number exists within [b]Q[/b]. To complete [b]Q[/b], one must augment it with additional numbers, one of which has this property.

A subspace S of a metric space M is said to be [b]closed[/b] if every convergent (in M) sequence of s[TEX]\in[/TEX]S converges to a limit in S. The closure of an (arbitrary) subspace of M is that subspace augmented with the limits of all its convergent (in M) subsequences.

Example: Let S = {a[sub]k[/sub], b[sub]k[/sub]} for all k[TEX]\ge[/TEX]0, with a[sub]k[/sub] and b[sub]k[/sub] defined as above. Viewed as a subspace of [b]Q[/b], S is already closed. (the only sequences within S convergent within [b]Q[/b] are those which repeat the same value from some point onward. All other "convergent" sequences "converge" on [TEX]\sqrt{2}[/TEX], that is to say don't converge at all.) The closure of S is therefore S itself. Viewed as a subspace of the real numbers, the closure of S is S [TEX]\cup[/TEX] {[TEX]\sqrt{2}[/TEX]}.

SM88, you are not expected to understand any of this.

[QUOTE=science_man_88;249938]No, because the term pairwise disjoint implies the element s[TEX]\in[/TEX] S not be part of more than one element p[TEX]\in P[/TEX] as the partition is like a collection of subsets . and pairwise disjoint implies they don't overlap.[/QUOTE]

Perfect.

[QUOTE]unless by definition the word set implies more than one element then yes it is possible that a element s\in S could also be a subset. (though a rather small one)[/QUOTE]

"Set" does not imply more than one element, and you have already met sets with just one element, as well as one with no elements at all.

Please give an example of a set with an element which is also a subset. Hint: What subset does every set have?

[QUOTE=Mr. P-1;249942]Perfect.



"Set" does not imply more than one element, and you have already met sets with just one element, as well as one with no elements at all.

Please give an example of a set with an element which is also a subset. Hint: What subset does every set have?[/QUOTE]

[TEX]\empty[/TEX]