union/intersection of family of sets - some basic question

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porky_...@my-deja.com

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May 8, 2005, 8:12:29 PM5/8/05
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I'm trying to understand the union and intersection of indexed families
of sets, and got a bit confused with notation. Hopefully somebody can
clarify it. I understand the notation

\bigcup {A}

or at least I believe that I do. The family in question consists of
only a single member A. It should be fairly straightforward to show
that

\bigcup {A} = A

How about the following:

\bigcup A

Here A is a set rather than a family of sets. So why would we want an
expression like this in a first place. Yet apparently it's perfectly
valid one. Would appreciate any comments regarding this notation. As I
understand, the families of sets and indexing is extremely important if
you want to move to more advanced topics (e.g. point-set topology).
TIA.

FunnyGuy

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May 8, 2005, 8:41:37 PM5/8/05
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In ZF there is a *union axiom* stating something like:
For any set X there is a unique set Y such that y in Y
iff y in x for some x in X.
Thus \bigcup X can then be defined as the unique set Y such that y in Y
iff y in x for some x in X.

Furthermore the binary union operator,
\cup, can then be defined as follows:
X \cup Y == \bigcup {X,Y}

This is the binary union operator you are probably accustomed to.
Learn for yourself how the usual binary intersection operator can be
defined (there is no axiom of intersection AFAIK).

Anyway like i always say I normally get things wrong be default so I await
embarassing correction but I hope the gist of what i said is sound.

William Elliot

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May 8, 2005, 9:34:11 PM5/8/05
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On Sun, 8 May 2005, porky_...@my-deja.com wrote:

> I'm trying to understand the union and intersection of indexed families
> of sets, and got a bit confused with notation. Hopefully somebody can
> clarify it. I understand the notation
>
> \bigcup {A}
>
> or at least I believe that I do. The family in question consists of
> only a single member A. It should be fairly straightforward to show
> that
>
> \bigcup {A} = A
>

x in \/{A} iff some U in {A} with x in U
iff some U = A with x in U
iff x in A
\/{A} = A

> How about the following:
>
> \bigcup A
>

\/A = { x | some U in A with x in U }

> Here A is a set rather than a family of sets. So why would we want an
> expression like this in a first place. Yet apparently it's perfectly
> valid one. Would appreciate any comments regarding this notation.

Let tau be a topology for S, then \/tau = S.
C is a cover of S when C subset tau and S = \/C

> As I understand, the families of sets and indexing is extremely
> important if you want to move to more advanced topics (e.g. point-set
> topology).

Whoops, you know topology? If not, the for the above topological
examples, for any set S, a topology tau for S is a subset of P(S)
(where P(S) = { A | A subset S }) and some other properties of no
concern for the examples above.

Usually indexing isn't needed, it's just a crutch.

How about /\{A} = A and /\A ?
For example A = { [0,x) | x in R }; /\A = {0}; \/A = [0,oo)

Exercise: \/{ A,B } = A \/ B; /\{ A,B } = A /\ B

Exercise: \/nulset = nulset
Booby trap: /\nulset = ??

Norman Megill

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May 8, 2005, 9:55:56 PM5/8/05
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porky_...@my-deja.com wrote:

> I'm trying to understand the union and intersection of indexed families
> of sets, and got a bit confused with notation. Hopefully somebody can
> clarify it.

Be careful to distinguish the standard from the indexed union. The
latter usually has an index range as a subscript in textbooks.

The Metamath pages have a large collection of theorems about standard
and indexed union, and might be helpful as a supplement to other
material; you'll have to decide.

The standard (nonindexed) \bigcup is defined here:

http://us2.metamath.org:8888/mpegif/df-uni.html

Binary union is defined here:

http://us2.metamath.org:8888/mpegif/df-un.html

X \cup Y == \bigcup {X,Y} is here:

http://us2.metamath.org:8888/mpegif/unpr.html

\bigcup {A} = A is here:

http://us2.metamath.org:8888/mpegif/unisn.html

Indexed union is here. If you click on "Related theorems" you'll see
a large collection of results. Study the indexed union definition
carefully; the concept can be confusing.

http://us2.metamath.org:8888/mpegif/df-iun.html

The following theorem defines indexed \bigcup in terms of standard \bigcup:

http://us2.metamath.org:8888/mpegif/dfiun2.html

The following theorem defines standard \bigcup in terms of indexed \bigcup:

http://us2.metamath.org:8888/mpegif/uniiun.html

Hope some of this helps.

--
Norm Megill http://public.xdi.org/=nm

porky_...@my-deja.com

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May 8, 2005, 11:04:48 PM5/8/05
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Thanks to everyone who's responded and especially the links. I was
going to ask to point me to some resources. Are there textbooks with a
solid intro to set theory, and targeted to topology? The only topology
textbook I have is Bert Mendelson 'Intro....' and I'm getting Munkres.
Any other recommendations? TIA.

Stephen J. Herschkorn

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May 8, 2005, 11:49:25 PM5/8/05
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porky_...@my-deja.com wrote:

You should do fine with Munkres. If you want other sources, you can
look at Kelly's General Topology. I think *every* mathematician should
read Halmos, Naive Set Theory. It is a gem.

--
Stephen J. Herschkorn sjher...@netscape.net

Arturo Magidin

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May 9, 2005, 10:17:57 AM5/9/05
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In article <1115597549.0...@o13g2000cwo.googlegroups.com>,

porky_...@my-deja.com <porky_...@my-deja.com> wrote:
>I'm trying to understand the union and intersection of indexed families
>of sets, and got a bit confused with notation. Hopefully somebody can
>clarify it. I understand the notation
>
>\bigcup {A}
>
>or at least I believe that I do. The family in question consists of
>only a single member A. It should be fairly straightforward to show
>that
>
>\bigcup {A} = A
>
>How about the following:
>
>\bigcup A

Here you think of A as the family; the elements of A are the sets
whose union you are taking. This is sometimes called "the amalgamated
union of A" and is the union of all x such that x is an element of A.


--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
mag...@math.berkeley.edu

Arturo Magidin

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May 9, 2005, 3:00:31 PM5/9/05
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In article <1115664837.2...@f14g2000cwb.googlegroups.com>,
porky_...@my-deja.com <porky_...@my-deja.com> wrote:
>A follow-up question on ZF Union Axiom.
>
>Since several responders referred to Union Axiom, I've checked some
>resources on ZF set of axioms. I think I ''almost"" understand the
>Union Axiom except for one thing, hence I'm asking for some help once
>more.
>
>Suppose I have set x (all the resources I saw use the lower-case
>letters for either sets or members of sets so I'll do as well). The
>Union Axiom states that there is some corresponding set y such that
>t \in y iff
>t \in z and z \in x, where z is some set, and such set exists.
>This axiom is for any set x.
>
>OK, say we have
>x = { {1, 2}, {2, 3} }, and then of course we construct y = \bigcup x
>= { 1, 2, 3 } so
>with t=1, 1 \in {1, 2} and {1, 2} \in x. Here z = {1, 2}
>
>So far so good.

Yes.

>But the statement 't \in z and z \in x', at least the
>way I interpret it, implies that z is a set, and thus x is a family of
>sets.

Yes. In ZF, there are no ur-elements (elements which are not
sets). Everything is a set, and all sets are sets of sets.


>How about x = {1, 2, 3}?

In ZF, 0 is the empty set, 1 represents the set {0} = {emptyset}
(i.e., the set whose only element is the empty set), 2 represents the
set {0,1} = {emptyset, {emptyset}}, and
3 = {0,1,2} = {emptyset, {emptyset}, {emptyset, {emptyset}}}.

> The axiom states '\forall x', so it
>should work for x = {1, 2, 3},

If you are using ur-elements (elements that are not sets), then U x
would be the empty set. If you are interpreting 1, 2, and 3 as they
are usually interpreted in ZF, then y would be the set
{emptyset, {emptyset}, {emptyset,{emptyset}} = 3.

> right? But then in this context what
>would be the set z? Say, we've constructed y = {1, 2, 3}.

But you haven't. The axiom of union says that y should be the result
of taking 1 union 2 union 3, whatever 1, 2, and 3 are. If they are
sets, take their usual union. If they are ur-elements, then their
union is the empty set.

> Now let t =
>1.
>t \in y \iff t \in ??? and ??? \in x. So what is ???.

porky_...@my-deja.com

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May 9, 2005, 2:53:57 PM5/9/05
to
A follow-up question on ZF Union Axiom.

Since several responders referred to Union Axiom, I've checked some
resources on ZF set of axioms. I think I ''almost"" understand the
Union Axiom except for one thing, hence I'm asking for some help once
more.

Suppose I have set x (all the resources I saw use the lower-case
letters for either sets or members of sets so I'll do as well). The
Union Axiom states that there is some corresponding set y such that
t \in y iff
t \in z and z \in x, where z is some set, and such set exists.
This axiom is for any set x.

OK, say we have
x = { {1, 2}, {2, 3} }, and then of course we construct y = \bigcup x
= { 1, 2, 3 } so
with t=1, 1 \in {1, 2} and {1, 2} \in x. Here z = {1, 2}

So far so good. But the statement 't \in z and z \in x', at least the


way I interpret it, implies that z is a set, and thus x is a family of

sets. How about x = {1, 2, 3}? The axiom states '\forall x', so it
should work for x = {1, 2, 3}, right? But then in this context what
would be the set z? Say, we've constructed y = {1, 2, 3}. Now let t =


1.
t \in y \iff t \in ??? and ??? \in x. So what is ???.

TIA.

Stephen J. Herschkorn

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May 9, 2005, 3:03:38 PM5/9/05
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porky_...@my-deja.com wrote:

In ZF, everything is a set. For example, if you mean the natural
numbers, 0 = {}, 1 = {0},. 2 = {0, 1}, and 3 = {0, 1, 2}. So U{1,
2, 3} = 3. (More generally, the union of a finte set of ordinals is the
largest ordinal in the set.)

On the other hand, there are some set theories with ur-elements which
are not sets. I suppose, literally, if 1, 2, and 3 are not themselves
sets, then U{1, 2, 3} = {}.

porky_...@my-deja.com

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May 9, 2005, 3:30:54 PM5/9/05
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>> right? But then in this context what
>>would be the set z? Say, we've constructed y = {1, 2, 3}.

>But you haven't. The axiom of union says that y should be the result
>of taking 1 union 2 union 3, whatever 1, 2, and 3 are. If they are
>sets, take their usual union. If they are ur-elements, then their
>union is the empty set.

This make perfect sense --- but, if I just read the Union Axiom, it
states that \forall x \exists y, but it doesn't say anything about how
we construct y. At least the defintions I saw: If I recall correctly,
it simply states that

\forall x \exists y \forall t
{ t \in y \iff \exists z (t \in z \land z \in x) }

So to me looks like the axiom states the existance of y but doesn't say
anything, at least not explicitly how we're going to construct y. Would
you mind to comment on that point?

TIA.

Arturo Magidin

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May 9, 2005, 3:37:44 PM5/9/05
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In article <1115667054.1...@f14g2000cwb.googlegroups.com>,

porky_...@my-deja.com <porky_...@my-deja.com> wrote:
>>> right? But then in this context what
>>>would be the set z? Say, we've constructed y = {1, 2, 3}.
>
>>But you haven't. The axiom of union says that y should be the result
>>of taking 1 union 2 union 3, whatever 1, 2, and 3 are. If they are
>>sets, take their usual union. If they are ur-elements, then their
>>union is the empty set.
>
>This make perfect sense --- but, if I just read the Union Axiom, it
>states that \forall x \exists y, but it doesn't say anything about how
>we construct y.

Of course not. Because the axiom of union is not about CONSTRUCTING
anything. It merely says that there is SOME set with the property that
every element of some element of x will be an element of y.


>At least the defintions I saw: If I recall correctly,
>it simply states that
>
>\forall x \exists y \forall t
> { t \in y \iff \exists z (t \in z \land z \in x) }

This is a slightly stronger version (but equivalent in the presence of
the axiom of specification). It says that for every set x, there
exists a set y with the property that the elements of y are exactly
the things which are elements of some element of x.

>So to me looks like the axiom states the existance of y but doesn't say
>anything, at least not explicitly how we're going to construct y.

It does not say anything about construction; it doesn't have to. ZF is
not about constructing things. The axioms are not about how you
construct things. The axioms are about what properties the objects
have or do not have, and about what sets you can guarantee exist or
not.

Dave Seaman

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May 9, 2005, 4:47:16 PM5/9/05
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On Mon, 9 May 2005 19:37:44 +0000 (UTC), Arturo Magidin wrote:
> In article <1115667054.1...@f14g2000cwb.googlegroups.com>,
> porky_...@my-deja.com <porky_...@my-deja.com> wrote:
>>>> right? But then in this context what
>>>>would be the set z? Say, we've constructed y = {1, 2, 3}.
>>
>>>But you haven't. The axiom of union says that y should be the result
>>>of taking 1 union 2 union 3, whatever 1, 2, and 3 are. If they are
>>>sets, take their usual union. If they are ur-elements, then their
>>>union is the empty set.
>>
>>This make perfect sense --- but, if I just read the Union Axiom, it
>>states that \forall x \exists y, but it doesn't say anything about how
>>we construct y.

> Of course not. Because the axiom of union is not about CONSTRUCTING
> anything. It merely says that there is SOME set with the property that
> every element of some element of x will be an element of y.

If we knew how to construct something, we wouldn't need an axiom to say
it exists. In fact, that's the reason we have a union axiom but no
intersection axiom. The intersection of a nonempty family of sets can be
shown to exist (can be constructed) from the other axioms of ZF, but the
union can't.

Consider the power set axiom as another example. We know from Cantor's
theorem that P(N), the power set of the naturals, is uncountable and
therefore its members cannot be enumerated or described; yet, the axiom
says the set exists.

--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&bookid=228>

FunnyGuy

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May 9, 2005, 7:42:47 PM5/9/05
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On Mon, 09 May 2005 12:30:54 -0700, porky_...@my-deja.com wrote:

> So to me looks like the axiom states the existance of y but doesn't say
> anything, at least not explicitly how we're going to construct y. Would
> you mind to comment on that point?

All I think of to say is "Welcome to the world of axiomatic set theory".

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