So you are saying that aleph_0 is not a member of the set of natural numbers yet the number of its members is aleph_0. I am however puzzled how all the members of the natural number set are finite yet it has aleph_0 members. — MoK

Let's look at some simpler examples.

1) Consider the set $\{3, 18, 58, 334\}$. That's a set with 4 elements, yet 4 is NOT an element of the set.

2) Consider the set $\{4, 18, 58, 334\}$. That's a set with 4 elements, and 4 IS an element of the set.

So we see that the cardinal number of a set MAY or MAY NOT happen to be an element of the set.

So far so good? Ok now into the infinite realm.

3) The set $\{1, 2, 3, ...\}$ does NOT contain the cardinal $\aleph_0$. I hope you can see that. If you claim otherwise, which element is it? It's not 1, it's not 2, it's not 3, etc.

4) The set $\{1, 2, 3, ..., \aleph_0 \}$ DOES happen to contain $\aleph_0$.

We see that in general, a set MAY or MAY NOT happen to contain its cardinality as a member.

In the case of the natural numbers, they do not contain their cardinality $\aleph_0$.

But there are surely $\aleph_0$ natural numbers, literally by definition. As we've seen from these examples, that fact gives us no information as to whether $\aleph_0$ is a member of the set of natural numbers. We have to look. And in this case, $\aleph_0 \notin \mathbb N$.

But if we tossed $\aleph_0$ into our set to get $\{1, 2, 3, ..., \aleph_0\}$, then we would have a set of cardinality $\aleph_0$ that happens to also have $\aleph_0$ as an element.

It can go either way, as the examples show.

By the way, how do we "toss $\aleph_0$ into the set?" The formal operation is taking a union. That is:

$\aleph_0 \notin \mathbb N$

but

$\aleph_0 \in \mathbb N \cup \{\aleph_0\}$

and BOTH sets have cardinality $\aleph_0$.

I defined the domain D that has this specific property, the number of its members, n, is a member as well. — MoK

Fine. But that's a different domain than the one I used.

Let me ask you this question: Are all members of the natural number set finite? — MoK

Yes, it's a common fact.

how the number of its members could be aleph_0 — MoK

The set of natural numbers is aleph_0. Every set is one-to-one with itself. So the set of natural numbers is one-to-one with alelph_0, so the cardinality of the set of natural numbers is aleph_0.

I am however puzzled how all the members of the natural number set are finite yet it has aleph_0 members. — MoK

There is no contradiction between:

Every member of the set of natural numbers is finite
and
The set of natural numbers has aleph_0 members

There is no contradiction between:

Every member of the set of natural numbers is finite
and
The set of natural numbers is not finite

It seems you have this false premise in your head:

If every member of a set S has property P then S has property P.

That premise is wrong and ridiculous. Get rid of it.

I should not be explaining basic logic to you. Your best bet would be to study a book on the first order predicate calculus and then one on set theory.

I am however puzzled how all the members of the natural number set are finite yet it has aleph_0 members. — MoK

Semantic puzzlement at the fact that one cannot finitely bound the natural numbers is understandable, even when having no syntactical puzzlement with regards to a formal definition of the naturals.

Firstly, there is presumably no semantic puzzlement about the situation in which one constructs the naturals one by one by counting upwards from zero, for in that case one assumes full control of the number generation process that one never finishes, for which at any time one has only constructed naturals that are a constructively finite distance from zero.

So i think that semantic puzzlement is in relation to arbitrarily large natural numbers that one assumes to exist but which aren't constructively defined.

For example, consider a guessing game between Alice and Bob, in which Alice privately thinks of a natural number which Bob then has to guess. Suppose that Bob is given an unlimited number of chances for guessing Alice's number. If Alice doesn't cheat by changing her number in response to Bob's guesses, then Bob has a winning strategy that will eventually terminate, such as guessing each natural number in turn by counting upwards from zero. But then suppose Alice does cheat to ensure that Bob will lose. How could Bob ever know it? Alice could for example secretly define her number to be one greater than Bob's largest guess. In which case her natural number isn't constructively finite in relation to Bob's strategy, in spite of being constructively finite in relation to Alice's strategy in the situation where Bob eventually gives up.

The poster didn't express puzzlement about the fact that there is no finite upper bound to the set of natural numbers.

D) By continuum I mean a set of distinct points without an abrupt change or gap between points.

A) Assume that continuum exists (assume that D is true)
P1) There is however either a gap between all pairs of points of the continuum or there is no gap
P2) We are dealing with the same point of the continuum if there is no gap between a pair of points
C1) Therefore there is a gap between all pairs of distinct points of the continuum (from P1 and P2)
C2) Therefore, the continuum does not exist (from A and C1) — MoK

Are you suggesting this proves real numbers are logically impossible, or are you arguing that there is no valid 1:1 mapping from the set of real numbers to the actual world? I ask, because it fails to do the former.

I am looking for proof that the set of natural numbers that each its member is finite has aleph_0 members.

The set of natural numbers is aleph_0. — TonesInDeepFreeze

What do you mean by this? Do you mean that the set of natural numbers is the set of aleph_0? aleph_0 is a number. How could you treat it as a set?

Every set is one-to-one with itself. So the set of natural numbers is one-to-one with alelph_0, so the cardinality of the set of natural numbers is aleph_0. — TonesInDeepFreeze

I don't understand this argument. How could aleph_0 be a number and a set at the same time?

Please accept my sincere apology for not replying to your post earlier. I read your post a few times in the past and had a difficult time understanding it. Unfortunately, I forgot to respond to your post. I recalled forgetting to reply to this post when I saw your second post in my thread. I read this post more than ten times trying to understand what you are trying to say but I have to say I am still ignorant in many parts. Please find my reply in the following.

Formally, the classical continuum "exists" in the sense that that it is possible to axiomatically define connected and compact sets of dimensionless points that possesses a model that is unique up to isomorphism thanks to the categoricity of second order logic. — sime

So classical continuum exists. I had to read about second-order logic, first-order logic, and zeroth-order logic trying to make sense of what you are trying to say here. Unfortunately, I don't understand what you are trying to say in the bolded part, the last part of the paragraph. Do you mind elaborating?

But the definition isn't constructive and is extensionally unintelligible for some of the reasons you pointed out in the OP. — sime

Do you mind elaborating on what problem you specifically have in your mind?

Notably, Dedekind didn't believe in the reality of cuts of the continuum at irrational numbers and only in the completeness of the uninterpreted formal definition of a cut. — sime

Do you by cut mean the exact position of an irrational number for example? What do you mean by the bolded part?

Furthermore, Weyl, Brouwer, Poincare and Peirce all objected to discrete conceptions of the continuum that attempted to derive continuity from discreteness. For those mathematicians and philosophers, the meaning of "continuum" cannot be represented by the modern definition that is in terms of connected and compact sets of dimensionless points. — sime

Ok, thanks for the reference.

E.g, Peirce thought that there shouldn't be an upper bound on the number of points that a continuum can be said to divide into, — sime

My understanding is that there is no upper bound on the number of points that a continuum can be divided into. I however don't understand whether he agrees with the classical notion of continuum or not. If not, what is his point?

whereas for Brouwer the continuum referred not to a set of ideal points, but to a linearly ordered set of potentially infinite but empirically meaningful choice sequences that can never be finished. — sime

Do you mind discussing his results further?

The classical continuum is unredeemable, in that weakening the definition of the reals to allow infinitesimals by removing the second-order least-upper bound principle, does not help if the underlying first-order logic remains classical, since it leads to the same paradoxes of continuity appearing at the level of infinitesimals, resulting in the need for infinitesimal infinitesimals and so on, ad infinitum.... whatever model of the axioms is chosen. — sime

So, one cannot define infinitesimal in the classical continuum. Is that what you are trying to say?

Alternatively, allowing points to have positions that are undecidable, resolves, or rather dissolves, the problem of 'gaps' existing between dimensionless points, in that it is no longer generally the case that points are either separated or not separated, meaning that most of the constructively valid cuts of the continuum occur at imprecise locations for which meta-mathematical extensional antimonies cannot be derived. — sime

What do you mean by antimony here?

Nevertheless this constructively valid subset of the classical continuum remains extensionally uninterpretable, for when cut at any location with a decidable value, we still end up with a standard Dedekind Cut such as (-Inf,0) | [0,Inf) , in which all and only the real numbers less than 0 belong to the left fragment, and with all and only the real numbers equal or greater than 0 belonging to the right fragment, which illustrates that a decidable cut isn't located at any real valued position on the continuum. Ultimately it is this inability of the classical continuum to represent the location of a decidable cut, that is referred to when saying that the volume of a point has "Lebesgue measure zero". And so it is tempting to introduce infinitesimals so that points can have infinitesimal non-zero volume, with their associated cuts located infinitesimally close to the location of a real number. — sime

So you are trying to say that introducing infinitesimal can resolve the problem of cut which is problematic for classical continuum.

The cheapest way to allow new locations for cuts is to axiomatize a new infinitesimal directly, that is defined to be non-zero but smaller in magnitude than every real number and whose square equals 0, as is done in smooth infinitesimal analysis, whose resulting continuum behaves much nicer than the classical continuum for purposes of analysis, even if the infinitesimal isn't extensionally meaningful. The resulting smooth continuum at least enforces that every function and its derivatives at every order is continuous, meaning that the continuum is geometrically much better behaved than the classical continuum that allows pathological functions on its domain that are discontinuous, as well as being geometrically better behaved than Brouwer's intuitionistic continuum that in any case is only supposed to be a model of temporal intuition rather than of spatial intuition, which only enforces functions to have uniform continuity. — sime

What do you mean by temporal and spatial intuition here?

The most straightforward way of getting an extensionally meaningful continuum such as a one dimensional line, is to define it directly in terms of a point-free topology, in an analogous manner to Dedekind's approach, but without demanding that it has enough cuts to be a model of the classical continuum. — sime

I am reading about point-free topology right now so I will comment on this part when I figure out what you mean with point-free topology.

E.g, one can simply define a "line" as referring to a filter, so as to ensure that a line can never be divided an absolutely infinite number of times into lines of zero length, and conversely, one can define a collection of "points" as referring to an ideal, so as to ensure that a union of points can never be grown for an absolutely infinite amount of time into having a volume equaling that of the smallest line. This way, lines and points can be kept apart without either being definable in terms of the other, so that one never arrives at the antimonies you raised above. — sime

I did read your links partially but I couldn't figure out what filter and ideal are. I don't understand how they resolve the problem of the classical continuum too but I buy your words on it.

Are you suggesting this proves real numbers are logically impossible, or are you arguing that there is no valid 1:1 mapping from the set of real numbers to the actual world? — Relativist

I directly attacked the continuum in the mathematical sense. The discussion is ongoing but it seems that the classical continuum exists but suffers from problems which are discussed here. The post is very technical and I have problems understanding it though.

I ask, because it fails to do the former. — Relativist

Do you mean later (instead of former)? If yes, it would be nice of you to elaborate.

I am looking for proof that the set of natural numbers that each its member is finite has aleph_0 members. — MoK

That's literally the definition of $\aleph_0$.

A set is defined to have cardinality $\aleph_0$ if the set can be placed into bijective correspondence with the natural numbers. Clearly the natural numbers themselves have this property, with the identity map as the bijection.

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

I don't understand this argument. How could aleph_0 be a number and a set at the same time? — MoK

All numbers are sets. In math, everything is a set.

What do you mean by this? Do you mean that the set of natural numbers is the set of aleph_0? aleph_0 is a number. How could you treat it as a set? — MoK

As a set, $\aleph_0 = \mathbb N = \{0, 1, 2, 3, 4, 5, ...\}$

I am aware of that. To avoid confusion, assume that the cardinality of the set of real numbers is X. How could one show that X is the least infinity namely aleph_0?

Sorry if I wasn't clear. I don't think you've actually proven real numbers are logically impossible. I haven't read through all the posts, so I don't know if this has already come out.

In real number theory, there are necessarily an infinite number of points between any two distinct points. This means it's logically impossible for there to be adjacent points. So when you say, "Therefore there is a gap between all pairs of distinct points of the continuum", is misleading because the "gap" contains infinitely many points. It's not a gap (which I think connotes an absence of points), it's an interval.

I am aware of that. To avoid confusion, assume that the cardinality of the set of real numbers is X. How could one show that X is the least infinity namely aleph_0? — MoK

It's not. Recall the Cantor diagonal argument, which shows that the cardinality of the reals is strictly greater than the cardinality of the naturals.

It's easy to prove that $|\mathbb R | = 2^{\aleph_0}$ and that this is strictly greater than $\aleph_0$.

The question of whether that happens to be equal to $\aleph_1$ or not is called the Continuum hypothesis. It's known to be independent of the standard axioms of math. That is, we can't prove it's true, and we can't prove it's false.

One attempt to resolve the issue is to try to find new, naturalistic axioms that settle the matter one way or another. Others argue that it's not even a well-defined question. The problem has been one of the driving forces in modern set theory ever since Cantor first posed it in 1878.

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

Ok, that I agree. How about whether the real number represents reality or not? Do you know any references that claim that reality is not continuous? I am currently reading a manuscript about loop quantum gravity and discrete time. I am a condensed matter physicist but I studied particle physics and cosmology to a good depth. I am somehow ignorant of loop quantum gravity and string theory though.

A mistake on my part and I am sorry for that. I should have written: "To avoid confusion, assume that the cardinality of the set of natural (I wrote real instead of natural) numbers is X. How could one show that X is the least infinity namely aleph_0?"

No, I haven't seen a general proof that the continuum can't map to reality. I've only seen arguments against specific mappings.

Does it matter? In terms of measurement, there a limit to how accurately we could possibly measure (at least, AFAIK)- so reality is discrete for all practical purposes.

A mistake on my part and I am sorry for that. I should have written: "To avoid confusion, assume that the cardinality of the set of natural (I wrote real instead of natural) numbers is X. How could one show that X is the least infinity namely aleph_0?" — MoK

https://math.stackexchange.com/questions/517040/proof-that-aleph-null-is-the-smallest-transfinite-number

I didn't ask for a proof that shows that aleph_0 is the least infinity but to show that X is the least infinity namely aleph_0.

I didn't ask for a proof that shows that aleph_0 is the least infinity but to show that X is the least infinity namely aleph_0. — MoK

A set is defined to have cardinality $\aleph_0$ if it can be bijected to the natural numbers.

Clearly the natural numbers can be bijected to the natural numbers, via the identity map; that is, the function that maps each natural number to itself.

If X is the cardinality of the natural numbers, X is $\aleph_0$ by definition.

Your questions arise only because you don't know anything about the subject, and also because (contrary to your claims) you don't carefully read the answers given you. And when you ask question Q1 and get an answer A1, you then have to ask Q2 to understand Q1, on and on. The way to break that regress is to study the subject from page 1 in forward, not reverse, order.

You need a book on the first order predicate calculus, then one on set theory. Nothing else will give you a proper understanding of the subject.

But until perhaps I get some moments to address your latest questions, I'll at least address this one (the answer is trivial, except you don't know anything about the subject to see that it's trivial):

the least infinity namely aleph_0. — MoK

"the least infinity" is not defined by you.

But we prove that there is no infinite cardinal less than aleph_0. Moreover, aleph_0 = the unique infinite cardinal K such that there is no infinite cardinal less than K.

Dfn. if x and y are ordinals, then x is less than y if and only if x is a member of y

Dfn. x is successor-inductive if and only if (0 is in x & for all n, if n is in x then xu{x})

Axm. There exists a successor-inductive set

Thm. There exists a unique set that is a subset of all successor-inductive sets
Proof: Put a nickel in my proof jukebox and I'll supply the proof

Dfn. w = the unique set that is a subset of all successor-inductive sets

Dfn. n is finite if and only if [put a nickel in my definition jukebox and I'll supply the definiens]

Dfn. n is an ordinal if and only if [put a nickel in the machine]

Dfn. n is a natural number if and only if (n is finite & n is an ordinal)

Thm. w = {n | n is a natural number}
Proof: Put a nickel in the machine

Thm. w is successor-inductive
Proof: Put a nickel in the machine

Thm. No ordinal is a member of itself
Proof: Put a nickel in the machine

Dfn. k is a cardinal if and only if (k is an ordinal & k is not one-to-one with a lesser ordinal)

Dfn. x is infinite if and only if x is not finite

Thm. every successor-inductive set is infinite
Proof: Put a nickel in the machine

Thm. Every cardinal is an ordinal
Proof: A cardinal is an ordinal that is not one-to-one with a lesser ordinal. So every cardinal is an ordinal.

Dfn. aleph_0 = w

Thm. w is infinite
Proof: w is a successor-inductive set

Thm. w is an ordinal
Proof: Put a nickel in the machine

Thm. There is no infinite cardinal less than aleph_0
Proof: aleph_0 is infinite. Every member of aleph_0 is finite. So no member of aleph_0 is infinite. So there is no infinite ordinal that is a member of aleph_0. So no infinite cardinal is less than aleph_0.

Thm. If K and L are cardinals, then exactly one of these: (1) K dominates L or (2) L dominates K or (3) K = L
Proof: K and L are ordinals. Put a nickel in the machine for proof that this trichotomy obtains for ordinals (axiom of choice not needed here)

Thm. aleph_0 = the unique infinite cardinal K such that there is no infinite cardinal less than K
Proof: Suppose L not= aleph_0 is an infinite cardinal such that there is no infinite cardinal less than L. So aleph_0 is not less than L. And L is not less than aleph_0. Contradicts previous theorem.

I am looking for proof that the set of natural numbers that each its member is finite has aleph_0 members. — MoK

fishfry has been giving you the info.

But to understand very well, you need to go in forward direction from page 1 of a book - from the simplest concepts to the more involved concepts that depend on the prior concepts - instead of foolishly trying to work backwards. Anyway:

Thm: If k is a cardinal and j is a cardinal less than k, then j and k are not one-to-one
Proof: By definition of 'is a cardinal', j is not one-to one with k

Thm: If S is one-to-one with a cardinal, then there is a unique cardinal such that S is one-to-one with it
Proof: See above theorem

Dfn. If S is one-to-one with a cardinal, then card(S) = the unique cardinal that S is one-to-one with

Thm. if x is infinite and y is finite, then x and y are not one-to-one
Proof: Put a nickel in the machine

Thm. w is not one-to-one with any member of w
Proof: w is infinite but every member of w is finite

Thm. w = {n | n is a natural number} is a cardinal
Proof: w is an ordinal and w is not one-to-one with any member of w, so w is not one-to-one with any ordinal less than w

Thm. x is one-to-one with x
Proof: the identity function on x is an injection from x onto x

Thm. card({n | n is a natural number}) = aleph_0
Proof: {n | n is a natural number}) = aleph_0. aleph_0 is one-to-one with aleph_0

Do you mean that the set of natural numbers is the set of aleph_0? — MoK

Yes. I said it many posts ago in a post you claimed to have read.

aleph_0 is a number. — MoK

Set theory doesn't provide a definition of 'is a number'. Rather, set theory provides individual definitions of such things as 'is a natural number', 'is a real number', 'is an ordinal number', 'is a cardinal number'. In those, 'number' is not taken as a standalone adjective. We could just as well have this terminology in instead: 'is a natnum', is a 'realnum', 'is an ordinalnum', 'is a cardnum'. We don't make proofs by recourse to ascribing any properties to being "a number" in and of itself. Also, instead of 'ordinal number' and 'cardinal number' we may as well just say 'ordinal' and 'cardinal'.

Anyway, aleph_0 is an ordinal and it is the least infinite ordinal, and it is a cardinal and it is the least infinite cardinal.

How could you treat it as a set? — MoK

Everything in set theory is a set. Natural numbers, real numbers, ordinals, cardinals ... are sets.

By definition aleph_0 = w, and we prove that w = {n | n is a natural number}.

I mentioned that previously.

I'm keeping to first order set theory. The point is that in set theory, any two complete order fields are isomorphic. You can think of that as saying that any two complete ordered fields are the same as one another except we changed the names of the elements. In other words, they are structurally the same.

/

d is a Dedekind cut if and only if (1) d is a non-empty proper subset of the set of rational numbers & (2) for all x, if x is a member of d and y<x, then y is a member of d & (3) d has no greatest member

/

one cannot define infinitesimal in the classical continuum — MoK

Be clear. There is a difference between (1) defining the adjective 'is an infinitesimal' and (2) proving the existence of a particular infinitesimal and naming it.

In set theory in which we formulate the system of real numbers, we can define 'is an infinitesimal', and prove that no real number is an infinitesimal, just as I have in this thread. And in set theory we can prove that there is a system with hyperreals (including infinitesimals) and prove that such a system is not isomorphic with the system of real numbers but that that there is a subsystem of hyperreals that is isomorphic with the system of real numbers. However, in plain set theory, even though we can prove the existence of a system of hyperreals, at least the usual methods (compactness theorem or ultrafilters) don't define a particular system of hyperreals.

/

A filter is a certain kind of set. an ultrafilter is a certain kind of filter. ultrafilters are used to prove the existence of a system of hyperreals.

It is folly for you to be trying to figure out what ultrafilters are and how they play into hyperreals when you don't even know page 1 of set theory.

Therefore there is a gap between all pairs of distinct points of the continuum" — Relativist

It turned out that by 'gap' @MoK meant 'interval'. His thinking is hopelessly confused.

I brought that to his attention and he accepted it. The residual question is: does the continuum map into anything in the real world - in any meaningful way.

I don't opine here on that other question. But what do you mean by "maps into"? Do you mean "there is a function into" or do you mean "there is a one-to-one function into"? I surmise you mean the latter.

Yes, I mean a 1:1 relation between real numbers and something that exists. But also more than a trivial assertion (e.g. "there's infinitely many spatial points in a 1 foot length") Supertasks have been used to show there isn't such a mapping for some cases.

Supertasks have been used to show there isn't such a mapping for some cases. — Relativist

Zeno's paradox? Other? What arguments are you referring to that there is no injection from the set of real numbers into "the world"?

What arguments are you referring to that there is no injection from the set of real numbers into "the world"? — TonesInDeepFreeze

Here's my (meta-)argument.

If the real numbers are instantiated in the real world, then questions such as the axiom of choice and the Continuum hypothesis become subject to physical experiment.

Since no physics postdoc has ever applied for a grant to study such matters, I conclude that no physicist takes these questions seriously; and that in any event, such questions are so far beyond experimental investigation as to be meaningless.

Let me put it another way. If the real numbers have representation in the physical world, do you think there are physical nonmeasurable sets? If so, does that make the Banach-Tarski theorem a fact of the physical world?

Doesn't the absurdity of ever investigating such questions argue against the physical reality of the real numbers?

Supertasks have been used to show there isn't such a mapping for some cases. — Relativist

This I disagree with. Word games involving supertasks tell us nothing about physics.

First, again, I don't know what the poster means by "the real world" so I don't know what firm and clear notion there is of an injection from the set of real numbers into "the real world".

Also, the argument "There are no relevant experiments regarding surrounding aspects of the reals, therefore there is no such injection" requires the premise, "If there is such an injection, then there are relevant experiments regarding surrounding aspects of the reals". But how would we rule out that there could be an injection but no relevant experiments regarding surrounding aspects of the reals, or that there could be an injection but no known relevant experiments regarding surrounding aspects of the reals?

If the real numbers are instantiated in the real world — fishfry

The question was about an injection. What is the definition of "instantiated in the real world"? Does it just mean that there is the range of an injection from the set of real numbers?

If the real numbers are instantiated in the real world, then questions such as the axiom of choice and the Continuum hypothesis become subject to physical experiment. — fishfry

I don't know that that is the case. Moreover, cutting back to the question of an injection, I don't know that that the lack of someone thinking up an experiment would entail that there is no injection.

Moreover, would entertaining that there is an injection from the set of natural numbers N into the real world entail that there must be some experiment to conduct?

no physics postdoc has ever applied for a grant to study such matters — fishfry

I don't know that that is true.

such questions are so far beyond experimental investigation as to be meaningless. — fishfry

That might be the case. Indeed, even the question alone of the existence of an injection from the set of real numbers into "the real world" doesn't seem to me to have, at least so far, been given a firm and clear meaning.

such questions are so far beyond experimental investigation as to be meaningless. — fishfry

That might be the case; I don't know. But I don't see that to entertain that there might be an injection entails that there must be an experiment to conduct. But again, the question of the existence of an injection from the set of real numbers into "the real world" doesn't seem to me to have, at least so far, been given a firm and clear meaning.

Banach-Tarski — fishfry

I wouldn't think that to entertain that there is an injection from the set of reals into the real world entails that there is a physical version of Banach-Tarski. But again, the notion of such an injection is not definite enough for me to have much of a view anyway (as well as I'm not prepared to discuss details of Banach-Tarski).

I surely don't have a strong opinion on the question of the existence of an injection from the set of real numbers into "the real world", but at least I would want to ponder whether the question is even even meaningful to either affirm or deny.