Your picture of all of this is much too woozy
— TonesInDeepFreeze

I am sorry if that is true — Gregory

Then you are sorry.

.

Instead of points one works with lattices of open sets. I don't see this as improving the intuitive understanding of continua. Continuity in elementary topological spaces rests upon the idea of connectedness. The topology of the reals is fairly well established, so maybe start by studying this. — jgill

I'm tempted to think of Dedekind cuts as a mathematics joke, in the sense that when open sets L and R are used to define a Dedekind cut L|R for an irrational number r, the generated closed set [r] is disjoint from both L and R, and yet their union equated with the continuum. As I understand it, this disjointed representation of the continuum is in semantic conflict with the continuum's connected topology , which is ultimately the cause of the continuum being empirically uninterpretable and practically useless in real life without abuse of notation.

I think the interesting thing about the open-sets of the extended continuum (with -Inf and +Inf introduced as end points), is that they can be interpreted as representing propositions, due to the fact that they form a distributive lattice with a top element (-Inf,+Inf), whose join operation is set union representing logical disjunction, and whose meet operation is set intersection representing logical disjunction, in spite of this lattice lacking a bottom element (since the empty set isn't an open set).

Likewise, the closed-sets of the extended continuum can be interpreted as representing negated propositions, due to the fact that the intersection of an open set (-Inf, x) with a closed-set [x,y] is the empty set representing falsity. More specifically, any point [x] represents a false proposition under this interpretation, i.e. p[x] := NOT { p(-Inf,x) OR p(x,+Inf)}, where p denotes a predicate that maps open sets to propositions of some sort. This interpretation refrains from asserting the existence of a point x for which p is true, but it doesn't deny the existence of such points either. (To deny the existence of such points is to go from a pointfree topology to a pointless topology).

In short, the open sets of the extended rational numbers can represent propositions derived by coinduction with respect to a 'top' proposition that is continuous in the sense that it isn't isomorphic to any union of propositions whose domains are disjoint. This top proposition is empirically meaningful. For example, we generally don't consider a priceless Ming vase to be the same after smashing it and gluing the pieces back together. By going point-free with our continuous topology, at least initially, allows us to consider "points" as being defects that are introduced when damaging a continuum to produce a non-continuum, such as in the destructive testing of a smooth object.

Without comment on the rest of your post, the very first claim is incorrect :

when open sets L and R are used to define a Dedekind cut L|R for an irrational number r, the generated closed set [r] is disjoint from both L and R, and yet their union equated with the continuum. — sime

With your letters:

The interval [r] is just {r}.

And L and R are sets of rationals only; no irrationals are members of L or R. The only irrational in Lu{r}uR is r. Lu{r}uR is not the set of real numbers.

You are flat out incorrect in claiming that Lu{r}uR is the set of real numbers.

With my letters ('Q' for the set of rationals, 'c' for the complement operation on sets of rationals, '<' for the less-than relation on rationals):

Definitions of 'Dedekind cut' differ in particulars, but one concise definition:

D is a Dedekind cut
if and only if
D is a non-empty proper subset of Q and
For all t and v, if t in D and v < t, then v in D and
D has no greatest member

Then the definition: x is a real number if and only if x is a Dedekind cut

So your irrational r is itself a Dedekind cut, call it 'D'. And your R is c(D).

So D u {r} u c(D) = D u {D} u c(D). The only irrational in d u {r} u c(D) = D u {D} u c(D) is r = D.

In plain words, you are flat out incorrect that the union of the lower cut with the one member interval and with the complement of the lower cut is the set of reals. Neither the lower cut nor its complement have any irrational members.

You conflated the cut (which is a set of rationals) with a real interval.

I agree with you, but i probably didn't make myself clear enough. I'm saying that if L| R is a Dedekind cut consisting of two open sets (as is the case when the cut defines an irrational number that isn't already contained in R), then the union of L, L|R and R is a disjoint partition of the continuum, which is semantically problematic in being disconnected (even if not "disconnected" according to the narrow topological definition of connectedness in terms of open sets only). The closed interval [r,r] is what I meant by writing [r].

I'm saying that if L| R is a Dedekind cut consisting of two open sets — sime

I believe you should review the definition of Dedekind cuts. First, they can't be open sets, since (as Tones pointed out) L and R are sets of rationals.

then the union of L, L|R and R is a disjoint partition of the continuum, — sime

That's not true either, since L and R are sets of rationals. I'm repeating myself, but you keep making the same error. And there IS no continuum (within set theory) until we have the set of ALL Dedekind cuts, not just a single one.

But you are making a larger conceptual error. We already know what the real numbers are. The real numbers "form the unique (up to an isomorphism) Dedekind-complete ordered field."

https://en.wikipedia.org/wiki/Real_number

That axiomatic definition is sufficient to satisfy philosophical notions of the continuum, as well as derive all other properties of the reals. Once you have an ordered field with the least upper bound property, you couldn't ask for anything more in terms of a continuum.

The purpose of Dedekind cuts is so that if someone says, "Oh yeah? How do we know there even IS such a thing?" we can show them the construction and say, "We can construct such an object within set theory." So we show them the construction once (or work through it in real analysis class once in our lives) and never think about it again. From now on we only need the axiomatic properties, having convinced ourselves that we could always construct such an object within set theory if challenged.

It's analogous to defining 0, 1, 2, 3, ... as 0 = {}, 1 = {0}, 2 = {0,1}, etc. As Benacerraf noted, since there are infinitely many possible encodings of the natural numbers within set theory, none of them can actually "be" the natural numbers. Rather, this particular construction demonstrates that if challenged, we could construct a model of the Peano axioms within set theory.

The moral of the story is that we should not conflate the set-theoretic construction of a thing, with the thing itself, which is generally defined axiomatically by its characterizing properties.

I believe you should review the definition of Dedekind cuts. First, they can't be open sets, since (as Tones pointed out) L and R are sets of rationals. — fishfry

Yes, and that's what i meant. To explain myself clearer, I meant L and R to refer to open sets of rationals together with the entire set of rationals representing +Inf and the empty set of rationals representing -Inf. I'm not sure why people might have jumped to a different conclusion.

It is right and necessary to point out as I think you are meaning to imply, that traditionally Dedekind cuts are understood as being objects derived from sets of rationals, in which the rationals are understood to be constructed, or simply to exist, prior to the creation of open sets of rationals, which are then used to define the cuts called "irrational numbers". That approach to understanding the reals is very "bottom up", and possibly in contradiction with Dedekind's own understanding of his cuts, which i suspect might have been "top down" (see the SEP for more discussion on his thoughts about the continuum in relation to actual infinity).

In my case, i am stressing the benefits of a "top down" approach, in which one uses lattice theory to define a lattice of abstract elements that is isomorphic to the open sets of the rationals extended by end points. The open sets of the rationals are only intended to serve as a model of this lattice, which is free to not assume the existence of points and other closed sets.

I meant L and R to refer to open sets of rationals — sime

Can you give an example of an open set of rationals? You've used the term several times. With respect to the usual topology on the reals, the rationals are not open nor are the intersections of the rationals with open sets of reals. So I'm a little confused on your meaning of an open set of rationals.

ps -- If you already had the reals, you could define open sets of rationals in terms of the subspace topology. But since we are trying to construct the reals, I'm not sure if that works. Perhaps you can define a topology on the rationals as being generated by open intervals of rationals. That might work. Is that what you intend?

I don't know what you mean by "isn't already contained in R". R is the complement of L in the rationals. No irrational number is in R...period. (By the way, I wish the letter 'R' weren't being used for the complement of L, since I'd like to use the letter 'R' for the set of real numbers.)

My definition is more streamlined than taking a Dedekind cut to be a partition; with my definition, a Dedekind cut is simply the lower part.

There are two different things: (1) Do you want to provide a definition of 'open' and 'closed' in the rationals? (2) Consider open and closed in the reals.

Regarding (2), the definitions are ('R' stands for the set of reals, and x, y, and j range over reals):

S is open in R
<->
S is a subset of R and
Ax(x in S -> Ey(y>0 & Aj(|x-j|<y -> j in S)))

S is closed in R
<->
c(S) is open in R

Then, if I did the math right (?), no nonempty set of rationals is open in R.

In my case, i am stressing the benefits of a "top down" approach, in which one uses lattice theory to define a lattice of abstract elements that is isomorphic to the open sets of the rationals extended by end points. — sime

It's well known that the powerset of any set forms a distributive lattice.

"The power set of a set, when ordered by inclusion, is always a complete atomic Boolean algebra, and every complete atomic Boolean algebra arises as the lattice of all subsets of some set. "

https://en.wikipedia.org/wiki/Power_set

From there, can you walk me through your idea of how to use this to construct a model of the reals that satisfies the least upper bound property?

Can you give me an example of an open set of rationals? — fishfry

If I'm not mistaken, no non-empty set of rationals is open in the reals.

I think it is good you are getting back into the discussion. Who knows what might come out of this thread? My only reservation - and ignore if you like - is to perhaps not bring up the Stern–Brocot tree. — jgill

I hadn't lost interest but I needed time to reflect and read. Actually, I still need a lot more time to do that but this forum is too hard to resist. Already my return to the forum has set me on a new reading trajectory so it's already a win. As for Stern-Brocot...no promises :razz:

Post removed

The argument that MoK gave involved the real numbers and their ordering, and real intervals, and his own confused notion of infinitesimals. He gave a definition of 'continuum' that sputtered. And he argued that the reals are not a continuum. His arguments were a morass. And given his personal definition of 'a continuum', he was refuted that the reals are not one. — TonesInDeepFreeze

I believe his arguments indirectly captured the spirit of the Zeno's Paradoxes. I believe Zeno and MoK communicated their ideas informally but nevertheless there's something profound about them. Anyway, I don't want to further defend MoK's position as I have enough on my own plate to defend my own position.

@TonesInDeepFreeze: Apologies if this isn't the best forum etiquette, but I'm genuinely curious about your background. Would you mind sharing a bit about your education and career?

If I'm not mistaken, no non-empty set of rationals is open in the reals. — TonesInDeepFreeze

Right, that was the point of my question to @sime. I did convince myself that if you take the rationals by themselves, you can define a topology by all the "open" intervals (p,q) with p and q rational. That would be a base for a topology on the rationals I believe. Not 100% sure but I think it's likely. If you define the open sets as all the finite intersections and arbitrary unions, you'd have a topology. That might be what sime means. Can't think of any other interpretation of an open set of rationals, since no nonempty set of rationals is open in the subspace topology of the reals as you note.

So then my next question was, given that topology on the rationals, how does sime propose to construct the reals. I suspect that in the end he'd have to reinvent Dedekind cuts.

Define 'continua'. Preferably a mathematical definition. And most preferably not free-floating, hand-waving verbiage. — TonesInDeepFreeze

I keep writing a response to this and then deleting it. I'll be back on this ..

I did convince myself that if you take the rationals by themselves, you can define a topology by all the "open" intervals (p,q) with p and q rationa — fishfry

Yes, usually it is inherited from the usual topology on the reals. But ignoring the non-reals seems OK. Looks like it is connected as well. But not a linear continuum since it doesn't have the LUB property. Rusty here I"m afraid.

Yes, usually it is inherited from the usual topology on the reals. But ignoring the non-reals seems OK. Looks like it is connected as well. But not a linear continuum since it doesn't have the LUB property. Rusty here I"m afraid. — jgill

Can't be connected. The two sets $A = \{q \in \mathbb Q : q^2 \lt 2\}$ and $B = \{q \in \mathbb Q : q^2 \gt 2\}$ are both open (proof in a moment) and disconnect $\mathbb Q$.

In other words I'm disconnecting $\mathbb Q$ at $\sqrt 2$.

Why are $A$ and $B$ open?

I noted earlier that the intervals $(p, q)$ with $p, q \in \mathbb Q$ are a base for the topology. To see that $A$ is open, we write $A = \bigcup (-\infty, q)$ for all $q$ such that $q^2 \lt 2$. The proof that $B$ is open is analogous.

Of course I see that I have to include the unbounded intervals $(-\infty, q)$ and $(p, \infty)$ in my base. (ps -- I don't need to, that's just a notational convenience).

ps --

Ah now I'm the rusty one. All of the intervals (p,q) are open in the subspace topology of the rationals relative to the reals. So I didn't need all this machinery, my sets A and B are already open in the subspace topology.

Ok. You are disconnecting Q at a point that does not exist in Q. Thought you were restricting all points to Q. Usual approach to this is to assume the underlying reals.

Ok. You are disconnecting Q at a point that does not exist in Q. Thought you were restricting all points to Q. Usual approach to this is to assume the underlying reals. — jgill

Am I misunderstanding you? The rationals are not connected in the subspace topology because you can partition them into two disjoint open sets: those rationals whose squares are less than 2, and those whose squares are greater than 2. Didn't mention any irrationals.

OK. I used to teach elementary point set topology occasionally but it has been over a quarter century ago. Best for me to avoid this discussion at my age.

As I understand from this conversation:

Let:

Q for the set of rationals
Q_o for the set of open subsets of Q
Q_i for the set of open intervals of Q

R for the set of reals
R_o for the set of open subsets of R
R_i for the set of open intervals of R

Then:

Th. Q_o is a topology on Q
Th. Q_i is a base for Q_o
Th. Q is disconnected per Q_o

Th. R_o is a topology on R
Th. R_i is a base for R_o
Th. R is connected per R_o

The takeaway: R is connected per R_o but Q is disconnected per Q_o

/

For reference, definitions:

U for the 1-ary union operation. P for the power set operation. E for the existential quantifier. A for the universal quantifier. e for member_of. 0 for the empty set.

Df.
T is a topology on X
<->
(T subset_of PX &
XeT &
AS(S subset_of T -> US e T) &
Akm({k m} subset_of T -> k/\m e T))

Th.
T is a topology on X -> 0 e T

Df.
EX T is a topology on X ->
(B is a base for T
<->
(B subset_of T &
As(seT -> EJ(J subset_of B & s=UJ))))

Df.
B is a base for a topology on X
<->
ET(T is a topology on X &
B is a base for T)

Df.
S is an open subset of Q
<->
(S subset_of Q &
Ax(xeS -> Ey(yeQ & y>0 & Aj((jeQ & |x-j|<y) -> jeS))))

Df.
S is an open interval of Q
<->
(S is an open subset of Q &
Axyz((xeS & yeS & x<z<y) -> zeS))

Df.
S is an open subset of R
<->
(S subset_of R &
Ax(xeS -> Ey(yeR & y>0 & Aj((jeR & |x-j|<y) -> jeS))))

Df.
S is an open interval of R
<->
(S is an open subset of R &
Axyz((xeS & yeS & x<z<y) -> zeS))

Df.
T is a topology on X ->
(X is disconnected per T
<->
EBC(BeT &
CeT &
B not= 0 &
C not= 0 &
B/\C=0 &
X=BuC))

Df.
T is a topology on X ->
(X is connected per T
<->
~ X is disconnected per T)

For a moment I was thinking q^2<2 normally is q<sqr(2) for positive q, but if irrationals do not exist this inequality is invalid. It seemed to disconnect at the sqr(2), which doesn't exist. Fuzzy thinking. Lets move on.

Twelve pages and I do not pretend to be able to follow all the math. A succinct report would be nice from anyone inclined to provide. Has it been established that the existence of the continuum is strictly a matter of definition (as I suppose), thus existing or not existing depending on definition within this-or-that context? Or does it demonstrably exist or not exist in some other, absolute, way.

Or is the discussion mostly about definitions themselves?

For a moment I was thinking q^2<2 normally is q<sqr(2) for positive q, but if irrationals do not exist this inequality is invalid. — jgill

The entire point of $q^2 \lt 2$ is to define that set without reference to irrationals.

In fact that set is the standard example of a set of rationals that's bounded above but has no least upper bound, showing that the rationals lack the least upper bound property.

Twelve pages and I do not pretend to be able to follow all the math. A succinct report would be nice from anyone inclined to provide. Has it been established that the existence of the continuum is strictly a matter of definition — tim wood

Correct, the real numbers are defined as the continuum. They can be proven to exist within set theory, but that has no bearing on what's true in the real world.

The OP's vague ideas about the real numbers were falsified and clarified early on.

The entire point of q2<2 is to define that set without reference to irrationals. — fishfry

You've made your point. Don't rub it in.

You've made your point. Don't rub it in. — jgill

Not my intention at all.

Define 'continua'. Preferably a mathematical definition. And most preferably not free-floating, hand-waving verbiage. — TonesInDeepFreeze

The following is not a formal, finalized definition, but I hope sets the stage for the discussion:

A continuum is a finite, continuous object with the potential for arbitrarily fine partitioning, characterized by the cardinality $2^{\aleph_0}$.

Finite object: Finite in the sense that its complete set of attributes can be fully described without invoking infinite processes.
Continuous object: In 1D, the proposed fundamental objects are of two types: (1) open-ended curves, which are inherently continuous, and (2) points. A composite 1D object is the union of these fundamental objects and is continuous if, when duplicates are removed, the following conditions are met:

• Points are connected to 0–2 curves (but not to other points).
• Curves are connected to 0–2 points (but not to other curves).
• No objects are disconnected from the composite structure.

Potential for arbitrarily fine partitioning: The continuum can be subdivided into an increasingly refined composite object made up of arbitrarily many fundamental elements, maintaining its continuity.
Characterized by the cardinality $2^{\aleph_0}$: The partitioning process can be described algorithmically, such that no algorithm can be devised allowing for further division. Although this algorithm would not halt if executed, the structure of the algorithm itself reveals that the potential for infinite subdivision aligns with the cardinality $2^{\aleph_0}$.

I have refrained from providing examples or illustrations for the sake of brevity, though they could help clarify my position.

https://thephilosophyforum.com/discussion/comment/934861

There has been unclarity in this discussion. Two concepts have not been held distinctly:

(1) the continuum (a noun)

(2) is a continuum (an adjective)

The use of those terms in mathematics is confusing to those who haven't read their definitions.

(1) I've given the definition of 'the continuum', though even that term itself is used in different ways:

(1a) the continuum = R

(1b) the continuum = <R L> where L is the standard ordering of R

(2) I mentioned that 'is a continuum' may mean different things in different contexts. But, at least in topology (which is central to this discussion), there is a clear definition. I'll give it here (hopefully, there are not many typos or mistakes):

Df.
T is a topology on X
<->
(T subset_of PX &
XeT &
AS(S subset_of T -> US e T) &
Akm({k m} subset_of T -> k/\m e T))

Df.
C is compact per X and T
<->
(T is a topology on X &
C = <X T> &
AS((S subset_of T & X = US) -> EF(F is finite & F subset_of S & X = UF)))

Df.
C is connected per X and T
<->
(T is a topology on X &
C = <X T> &
~EBC(BeT & CeT & B not= 0 & C not= 0 & B/\C=0 & X=BuC))

C is Hausdorff per X and T
<->
(T is a topology on X &
C = <X T> &
Apq(({p q} subset_of X & p not= q) -> EUV({U V} subset_of T & p e U & q e V and U/\V = 0)))

C is a continuum per X and T
<->
(C is compact per X and T &
C is connected per X and T &
& C is Hausdorff per X and T)

/

Th.
{S | S is an open subset of Q} is a topology on Q

Th.
{B | B is an open interval of Q} is a base for {S | S is an open subset of Q}

Th.
{S | S is an open subset of R} is a topology on R

Th.
{B | B is an open interval of R} is a base for {S | S is an open subset of R}

Show: ~ <Q {S | S is an open subset of Q}> is a continuum per Q and {S | S is an open subset of Q}

Show: <R {S | S is an open subset of R}> is a continuum per R and {S | S is an open subset of R}

/

Df.
(T is a topology on X & W is a topology on Y)
->
(f is an isomorphism between <X T> and <W Y>
<->
(f is 1-1 from X to W &
Aj(jeT <-> f(j) e W)))

I correctly mentioned that the system of real numbers is a complete ordered field and that famously it is a theorem that all complete ordered fields are isomorphic (applying the usual definition of 'isomorphic' for systems). But are all continuums isomorphic (applying the definition of 'isomorphic' above)?

/

This mathematics does not preclude other senses of 'is a continuum' that people may wish to specify in philosophy and science.