# Topology question

#### phoenixthoth

a line segment (including its endpoints) is toplogically equivalent to a point.

consider S a nonempty totally ordered (ie a set with a relation <= such that <= is reflexive, transitive, x<=y and y<=x imply x=y, and every pair is comparable--i think that's what total order means anyway) set.

i'm suspecting that the subsets of S of the form $$G_{y}:=\left\{ x\in S:y\leq x\right\}$$ form a topology on S. oh, let's assume S has a $$\leq$$-maximal element and minimal element to insure that S and Ø are in the topology. i think one may need zorn's lemma to prove that the union of open sets is open. for finite intersections of $$G_{y}$$'s, take the max of the y's, call that max y' and then consider $$G_{y'}$$ which is i believe the intersection of the others.

now is S equipped with this topology homeomorphic to a point (with the concrete topology)? is it obviously not homeomorphic?

well, if there is a universal set, then let S=U. S turns into a totally ordered set with <= meaning injection. S has a minimum and maximum <= element, namely Ø and U, and so U would be topologically equivalent to 1={Ø}. perhaps another equation up there with euler's formula would be:
{Ø}=1~U

note to Self: 435.

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a line segment (including its endpoints) is toplogically equivalent to a point.

No. A nondegenerate line segment has open sets in its interior, a point doesn't. A line segment is contractible to a point, but that is not the same as topological equivalence.

#### matt grime

Homework Helper
homotopic not homeomorphic, look 'em up.

#### phoenixthoth

ah, yes, homotopic. but a point has the same euler characteristic as a line segment, yes? my bad for thinking that meant they were homeomorphic and not just homotopic. sorry; it's been 6 years since i thought about topology.

i also realized that no function from S to {Ø} can have an inverse in most cases so they're obviously not homeomorphic.

the sorgenfrey line could look like this: $$G_{y}:=\left\{ x\in U:y<x\right\}$$ and then perhaps the sigma algebra generated by the collection $$\bigcup_{y\in U}G_{y}$$ could be the generalized interval concept organic is looking for which is to say that anything in this sigma algebra could be called a generalized interval.

then let ~ in the equation {Ø}=1~U mean 'is homotopic to'.

i suppose then the question is is U with the topology of either the "open" or "closed" sorgenfrey line homotopic to {Ø}=1?

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#### matt grime

Homework Helper
You cannot require your order to be induced by injections, because that you have sets with A<=B and B<=A and A not equal to B.

The usual partial ordering is by inclusion.

To attempt to get a total ordering would decree any two sets of the same cardinality are equal

A sigma algebra and a topology aren't the same, which do you want?

Besides, homotopy requires you to define contintuity and to have the cartesian product of two sets defined in your theory, which I couldn't comment on.

#### phoenixthoth

Originally posted by matt grime
You cannot require your order to be induced by injections, because that you have sets with A<=B and B<=A and A not equal to B.

The usual partial ordering is by inclusion.

To attempt to get a total ordering would decree any two sets of the same cardinality are equal

A sigma algebra and a topology aren't the same, which do you want?

Besides, homotopy requires you to define contintuity and to have the cartesian product of two sets defined in your theory, which I couldn't comment on.
the sigma algebra generated by the topology. ie the intersection of all sigma algebras containing the topology.

forget the universal set business for now if you will and just stick to a set with total ordering, a maximum and minimum element and the set {Ø}. *hmm* and 'index set' for the homotopy, eh? let me think about that. i want to say the index set is S itself.

i'm thinking if S was a lattice with three elements F<M<T how could one shrink {F,M,T} to {T} homotopically. i wish i had my copy of royden or my schaum's general topology! (well, that may be a frivolous interaction with 3 valued logic and a unitary logic.)

#### matt grime

Homework Helper
A homotopy (of topological spaces) X and Y are continuous maps f:X--->Y and g:Y--->X with the compotions homotpic to the resp identity maps.

A homotopy between maps r and s on some space X is a continuous map

h: XxI --->XxI

with h(x,0)=r(x) and h(x,1)=s(x)

and I is the unit interval (whcih I also don't know to be a set in your theory..)

I don't know if cartesian products exist in your universe (well power sets don't, so I dunno about these either).

dunno what you mean by index set.

As for lattices. do you mean a lattice in R^2 (or equivlently C, defined by two complex numbes x and z with z/x non-real)? Any two are homeomorphic.
(excepting degenerate cases).

#### phoenixthoth

Originally posted by matt grime
A homotopy (of topological spaces) X and Y are continuous maps f:X--->Y and g:Y--->X with the compotions homotpic to the resp identity maps.
thank you for the definition. i was wondering in what sense you mean homotopic to the resp. identity map.

A homotopy between maps r and s on some space X is a continuous map

h: XxI --->XxI

with h(x,0)=r(x) and h(x,1)=s(x)

and I is the unit interval (whcih I also don't know to be a set in your theory..)
that seems strange to me that a homotopy is between maps and not subsets of topological spaces. i'll have to digest that. the index set is the unit interval in this case and i'm wondering if that can be changed to any totally ordered set with a max and min. is it that if r restricted to some subset A of X and if s restricted to some subset B of X then A and B can be called homotopic to each other?
I don't know if cartesian products exist in your universe (well power sets don't, so I dunno about these either).
why wouldn't power sets exist? why wouldn't cartesian products exist?

As for lattices. do you mean a lattice in R^2 (or equivlently C, defined by two complex numbes x and z with z/x non-real)? Any two are homeomorphic.
(excepting degenerate cases)
i don't mean a lattice in R^2. i mean definition 2 in http://en.wikipedia.org/wiki/Lattice . so in that sense, is the lattice {F,M,T} with F<M<T (^ works like and and v works like or) homotpoic to {T}? if not, then the general totally ordered S is not homotopic to {Ø}.

#### matt grime

Homework Helper
Originally posted by phoenixthoth
that seems strange to me that a homotopy is between maps and not subsets of topological spaces. i'll have to digest that. the index set is the unit interval in this case and i'm wondering if that can be changed to any totally ordered set with a max and min. is it that if r restricted to some subset A of X and if s restricted to some subset B of X then A and B can be called homotopic to each other?
No, A and B aren't the homotopic things, whatever you think they might be - where did they come from? r and s are homotopic.

That is the defintion like it or not. It is about continuous deformations of maps. X is homotopic to Y if the identity map of X factors up to homotopy through Y. Note, it must be both ways. If you pick some other totally ordered 'index set' (that isn't what index set means normally) then what you have wouldn't necessarily be a homotopy, and you would have to prove something about generalizations. Homotopies arise because homotopic maps induce the same maps on chain complexes of the resolutions of spaces, where you might see another definition for homotopic maps. In particular homotopic spaces have the same homologies.

why wouldn't power sets exist? why wouldn't cartesian products exist?
Well, what is the power set of the universal set? There is a natural inclusion from one to the other, but no bijection so the power set is strictly larger than the largest set. Conclusion, the power set cannot satisfy the property of being a set. This is naive of course, you may have axioms that prevent this being true.

The existence of power sets is part of ZF.

Note, I'm not saying that no set has a power set but that at least one set's power set might be problematic.

i don't mean a lattice in R^2. i mean definition 2 in http://en.wikipedia.org/wiki/Lattice . so in that sense, is the lattice {F,M,T} with F<M<T (^ works like and and v works like or) homotpoic to {T}? if not, then the general totally ordered S is not homotopic to {Ø}.

So X would be the 3 element set with the open sets in the topology

{}, {1} {1,2} {1,2,3}

Y is the 1 point set with the trivial topology. {} {1}

f:X-->Y must send 1,2,3 all to 1

g:Y--> X sends 1 to..? Well, it could be any of them, the only requirement is that the inverse image of an open set is open.

Suppose it sends 1 to 1.

Is there a homotopy from Id_X to gf? fg is id_Y) Remember this must be in the product topology on XxI. I don't think so, because of the discrete nature of X: because the notional homotopy h(x,t)

satisfies h(x,1)=x, and varies continuously as t varies, it must be constant on x for all x, so Id is only homotopic to Id.

#### phoenixthoth

Originally posted by matt grime
No, A and B aren't the homotopic things, whatever you think they might be - where did they come from? r and s are homotopic.

That is the defintion like it or not. It is about continuous deformations of maps. X is homotopic to Y if the identity map of X factors up to homotopy through Y. Note, it must be both ways. If you pick some other totally ordered 'index set' (that isn't what index set means normally) then what you have wouldn't necessarily be a homotopy, and you would have to prove something about generalizations. Homotopies arise because homotopic maps induce the same maps on chain complexes of the resolutions of spaces, where you might see another definition for homotopic maps. In particular homotopic spaces have the same homologies.

Well, what is the power set of the universal set? There is a natural inclusion from one to the other, but no bijection so the power set is strictly larger than the largest set. Conclusion, the power set cannot satisfy the property of being a set. This is naive of course, you may have axioms that prevent this being true.

the subsets axiom was changed and the values of logic were increased from 2 to 3 to account for this. btw, even in that context, i proved that the universal set is the only set that equals (not just is in bijection with) its powerset. more generally, if a set x is mappable onto its powerset then it must be the universal set. the kernel of the idea was that the 'answer' to russell's paradox is 'maybe.'

So X would be the 3 element set with the open sets in the topology

{}, {1} {1,2} {1,2,3}

Y is the 1 point set with the trivial topology. {} {1}

f:X-->Y must send 1,2,3 all to 1

g:Y--> X sends 1 to..? Well, it could be any of them, the only requirement is that the inverse image of an open set is open.

Suppose it sends 1 to 1.

Is there a homotopy from Id_X to gf? fg is id_Y) Remember this must be in the product topology on XxI. I don't think so, because of the discrete nature of X: because the notional homotopy h(x,t)

satisfies h(x,1)=x, and varies continuously as t varies, it must be constant on x for all x, so Id is only homotopic to Id.
thanks. i see what you mean now about it being the maps that are homotopic. in complex analysis, i always viewed the curves as homotopic and not the functions that gave rise to those curves. still, i think there is sense to be made out of homotopic sets.

so what do you call it when it only satisfies half the definition of homotopy, if anything, if it even satisfies half?

#### matt grime

Homework Helper
You see why I was being careful about saying what set theoretic constructions I could assume you had?

I doubt there is any name for spaces where the are maps f,g with fg homotopic to id, and gf not, simply because that does not seem a very restrictive criterion. For the sake of argument, call it hemitopic.

the circle is hemitopic to any product of circles via the 'diagonal' map, x to (x,x,x...x), any topological space is hemitopic to any product of itself any numbre of times, infact. Moreoever, every set is hemitopic to a point. (As the relation of hemitopy is not symmetric, then this doesn't imply all spaces are hemitopic.) And therefore to any contractible space.

#### phoenixthoth

that while on one level seems to reduce it to a triviality on another level perhaps it is the mathematical equivalent of the philsophic notion of nonduality.

#### matt grime

Homework Helper
Originally posted by phoenixthoth
that while on one level seems to reduce it to a triviality on another level perhaps it is the mathematical equivalent of the philsophic notion of nonduality.
Erm, no. Would I regret asking what you mean by duality?

#### phoenixthoth

you might but this is certainly not the place. actually, canute is the best at explaining it. start a thread in philosophy: "what is duality/nonduality." i can pm you some links to authors who philosophize about it.

no, it's not trivial or no, it's not the mathematical equivalent of the philosophy of nonduality?

thank you for your insights, btw.

#### matt grime

Homework Helper
Originally posted by phoenixthoth
you might but this is certainly not the place. actually, canute is the best at explaining it. start a thread in philosophy: "what is duality/nonduality." i can pm you some links to authors who philosophize about it.

no, it's not trivial or no, it's not the mathematical equivalent of the philosophy of nonduality?

thank you for your insights, btw.

There are many dualities in mathematics, the ones I care about usally involve finding adjoints to functors cf Grothendieck Duality. Then there are Dual Spaces, not to mention Dual constructions of category theory.

We are not talking 'dual natures' such as wave particle.

#### phoenixthoth

isn't a vector space isomorphic to its dual's dual?

different in one sense but same in another sense, depending on perspective.

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#### matt grime

Homework Helper
For a finite dimensional vector space yes. That obviously fails in infinite dimensions. For, however for a module over a ring, there is more than just this.

In general a duality can be thought of as an exact pairing between spaces. Eg on an n-dimensional manifold there is a duality between H_r and H^{n-r} - a non-degenerate bilinear form.

I don't think it would help to explain the more advanced dualities, suffice to say that taking duals isn't as easy as people think because they only think in terms of finite dimensional vector spaces.

#### phoenixthoth

interesting. maybe you could expond on duality and how universal or not the statement, loosly phrased, "it 'equals' its double dual" is and what that would mean in semi-lay terms. i don't know that much about categories or opposites of categories or duality in that any sense.

not technically duality but what i meant by the last remark is what you probably knew but for others how two objects can be different, eg the sets {1,2,3} and {2,3,4}, yet the 'same' via an isomorphism. but even one ring isn't necessarily isomorphic to it with the same underlying set with a different multiplication rule!

it seems as though hemitopy can unify all sets in a perhaps trivial way (for it appears that saying 'all sets are sets' is a trivial unification/identification of a common property of all sets) in that if all sets are hemitopic to a point, then shouldn't all sets be hemitopic to each other?

#### matt grime

Homework Helper
Firstly the hemitopic statement is almost entirely vacuous as it is exactly equivalent to the statement 'a non-empty space is no-empty'.

Secondly, a space isomoprhic to its double dual is reflexive, it is well known and can be found in any reference on basic hilbert space theory/functional analysis.

l_p is dual to l_q iff 1/p+1/q = 1 for p and q non-zero.

l_0 has as its dual l_[infinity]

you don't want to know what the dual of l_[infinity] is, and I'm not sure I can eve describe it to be honest; it requires ultrafilters which behave oddly enough like the sigma algebras you mentioned (Moore-Smith convergeence and nets for reference).

if V is some module for a ring then V dual is space Hom_{foo}(V,something)

where the foo is what you want the maps to be equivariant wrt to and the something is some field/ring etc you map into.

example, the dual as you know it is

Hom_k(V,k)

the k-linear maps into the base field of a k-Vector space for some field k.

Note that for a vector space, V is not only isomorphic to V**, but to V* as well, it's just that the isomorphism is natural in the first case and unnatural in the second. Natural isos preserve arrows and unnatural ones reverse them

exampe a map V-->W gives maps

W*----.>V*

and

V**----> W**

This is what is meant by dual on category theory.

If F something defined by arrows and diagrams, then the dual construction of F is what you get when all the arrows are reversed.

example

product:

the product (categorical) of X and Y (when it exists) is an object XxY with maps XxY to X and Y with some universal property of commuting diagrams

dual construction, the coproduct is blah blah with maps from X and Y TO XuY satisfying some universal property of commuting diagrams with the arrows going the other way.

What you're calling duality isn't duality in any sense I'm aware of. What you are saying is that under the forgetful functor to the category SETS the objects become isomorphic.

#### curiousbystander

Originally posted by matt grime
Besides, homotopy requires you to define contintuity and to have the cartesian product of two sets defined in your theory, which I couldn't comment on. [/B]
Just my two cents before running off to class:
The power set
If A is a set the power set of A is (henceforth called P(A)) is also a set. It's defined to be the set of all subsets of A.
Topology-- loosely
A topology on a set is a collection of subsets that satisfy certain axioms (which I'll omit). We call those subsets open. You may think of the topology on a set A as a subset of P(A)... so when we're talking about a topological space A we're really talking about the set A, and the additional information encoded in some special subset of p(A). There may be more then one topology defined on the same set A-- in fact there are as many topologies on A as there are subsets of P(A) that satisfy the topology axioms.

Continuous Maps
Consider two topological spaces X and Y with topologies S and T.
There are many set-maps between X and Y. We have a simple criterion that allows us to decide if such a set map is continuous or not.

Any map that has the property: the preimage of ANY open set in Y is open in X, gets the title of continuous. We can restate this in the language of power sets as follows:

Any set map f:X->Y induces a map between p(Y) and p(X) called the preimage of f. This is less mysterious then it seems-- take any subset of the codomain (that's Y), and then the preimage of that set under f is subset of the domain (in this case X). That's just a map between p(B) and p(A).

We'll call it f^{-1} (sorry-- don't know how to do tex here). So a set map f:X->Y is continuous precisely when the induced set map f^{-1}:P(Y)->P(X)
has the property that f^{-t}(T) is contained within S.

The Cartesian Product
Let A and B be any two topological spaces, with topologies S and T.

A X B is also a topological space as we'll see below:

As a set A X B is the usual Cartesian product whose arbitrary elt is traditionally written as (a,b) where a is in a and b is in B. (I can get much fancier and more formal if circumstances warrant).

The topology on A X B is formed from
the topologies on A and B. It would be nice if S X T (which is almost a subset of P(A X B)) were a topology but it doesn't usually work that way. Let U be an elt of S (this means an open subset of A) and V an elt of T.
Define U X V to be an open set in A X B.
Now say that any subsets that can be built out of arbitrarily many unions and a finite number of intersections of these basic open subsets are also open...Toss in the empty set and all of A X B for good measure and voila-- topology on A X B. (why an arbitrary many unions and a finite number of intersections? That goes back to the topology axioms I omitted).

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#### matt grime

Homework Helper
I'm slightly curious, curiousbystander, as to why you've told me this. You should be aware that Phoenixthoth is not using ordinary set theory. And a priori there is no reason to assume that the product of two sets is a set. It turns out that his theory does allow one to take power sets, as it is one of the ZFC axioms which he includes, and the products of sets are I believe, sets.

Anyway, why the topology for dummies introduction? To define continuity? Notice the requirement for a universal set which can make things go whacky, and means we aren't dealing in ordinary maths, so all bets are off and it's better to be sure that things exist before assuming they are. In particular, as the power set of a universal set would be the universal set, I wouldn't be confident about claiming to define things in this space as obviously there are some counterintuitive results lyign around. For instance

UxU is categorically not equal to U for any set in ordinary set theory (except the empty set), yet it is in Phoenixthoth's. Sure you know what the product is and what the powersets are now?

#### NateTG

Homework Helper
Originally posted by phoenixthoth

consider S a nonempty totally ordered (ie a set with a relation <= such that <= is reflexive, transitive, x<=y and y<=x imply x=y, and every pair is comparable--i think that's what total order means anyway) set.

i'm suspecting that the subsets of S of the form $$G_{y}:=\left\{ x\in S:y\leq x\right\}$$ form a topology on S.
What you constructed is only a topology if you allow for $$y \notin S$$ and use some extended $$\leq$$, or you can show that $$S$$ is complete. Otherwise, if, for example, $$S = \mathbb{Q}$$ then $$\bigcup_{y^2 \leq 2} G_{y}$$ is a union of open sets, but is not in your list of open sets.

#### phoenixthoth

thanks, nate.

matt, why can't powersets exist in my theory?

or cartesian products? (well, i see that if powersets don't exist, then the usual construction of c.p.'s don't exist either, so this goes back to my first question.)

also, could you post your proof, not by contradiction, of the statement no set x can be mapped onto its powerset, under "the search for absolute infinity?" or a link to that proof?

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#### matt grime

Homework Helper
When i wrote that thing about power sets, i did not have a reference for your theory handy, so i did not know your axioms. now that i know that you are using ZFC+ something, I now know power sets exist. I do not at the moment know if that is consistent as i've not thought about it. I can give easy non-contradictory proofs that N is not in contradiction with P(N), I don't know if it translates into other sets.

I think it might do:

Given a set S, and a map f to P(S), define T in P(S) by t in T iff t not in f(t). then T is not in the image of f. Hence there is no bijection from S to P(S).

check that in your three valued logic.

#### phoenixthoth

the issue with that reasoning is that iff takes on a whole new meaning in three-valued logic. this is discussed in the truth table for A<->B at the beginning of page 2. in the + part of my theory, the logical connective extension of iff is what is represented by the circle and for crisp, binary entities, it means the same as iff. this aspect is discussed in the truth tables on page 3 and 4.

what i'm suspecting is that one may have to adopt an axiom that the power set of a set is larger than a set except in the case of the universal set; i believe that's an axiom i can live with.

the thing is i'm not well educated in the paradox of there being no largest ordinal. one way around that may be to say that U does not have an ordinal number.

anyways, the details of why this proof would fail to be a proof are, i believe in paragraph 3 on page 6. however, it dissolves the proof that a set cannot be mapped onto its powerset hence why i think that may have to be an axiom (except for the universal set).

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